The inverse function theorem and the reciprocal-slope picture behind it, differentiating a general inverse, the extension of the power rule to rational exponents, and the derivatives of all six inverse trigonometric functions obtained by a right-triangle argument — every one of them algebraic.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 3 — Derivatives
Derivatives of Inverse Functions
Objectives
Five outcomes. Every one is the chain rule applied to the statement that two functions undo each other.
OpenStax Calculus Volume 1, §3.7 Derivatives of Inverse Functions §3.7, pp. 259-267 — the section these objectives are drawn from
Warm-up
Section 1.4 defined an inverse by the two cancellation equations, and Section 3.6 supplied the chain rule. Putting them together settles this whole section.
Discussion prompt
Differentiate both sides of the equation saying that f of its own inverse returns the input. What comes out?
Hint: The left side is a composition, so the chain rule applies.
Answer:
\[ f\big(f^{-1}(x)\big) = x \]
\[ f'\big(f^{-1}(x)\big)\cdot\big(f^{-1}\big)'(x) = 1 \]
Solving for the inverse's derivative gives it immediately as a reciprocal. That single line is the inverse function theorem, and everything else in this section is an application of it.
Note where the chain rule was essential: without it the left side could not have been differentiated at all, since it is a composition rather than any arithmetic combination.
Concept
Reflecting a graph in the line y equals x swaps rise and run, so it turns a slope into its reciprocal. Differentiating the cancellation equation with the chain rule makes that precise, and it fixes where each derivative must be evaluated.
the inverse function theorem — If f is differentiable and one-to-one with a non-zero derivative, its inverse is differentiable and the inverse's derivative at x is one over f prime evaluated at the inverse's output at x.
\[ \big(f^{-1}\big)'(x) = \frac{1}{f'\big(f^{-1}(x)\big)} \]
The condition that f prime not vanish is doing real work. Where the original has a horizontal tangent, the reflection has a vertical one, and the inverse has no derivative there at all.
Figure (svg): The inverse function theorem, with the two evaluation points marked
OpenStax Calculus Volume 1, §3.7 Derivatives of Inverse Functions §3.7, pp. 259-261
Section
Section 1
Concept
A line of slope m reflected in the diagonal becomes a line of slope one over m, because reflecting swaps the roles of rise and run. Since a tangent line reflects to a tangent line, the inverse's slope is the reciprocal of the original's.
reciprocal slopes — At corresponding points, the slope of an inverse's graph is the reciprocal of the slope of the original's. The correspondence is the reflection in the line y equals x, which exchanges the two coordinates.
\[ (a, b) \text{ on } f \iff (b, a) \text{ on } f^{-1} \]
The picture makes the theorem plausible and the chain rule makes it a proof. Both are worth having: the picture tells you the answer must be a reciprocal, and the algebra tells you where to evaluate it.
Figure (svg): A function and its inverse reflected in the diagonal, with reciprocal slopes at matched points
OpenStax Calculus Volume 1, §3.7 Derivatives of Inverse Functions §3.7, pp. 259-262 — the inverse function theorem
Picture it
A function and its inverse with tangents at a matched pair.
Figure (svg): A function and its inverse reflected in the diagonal, with reciprocal slopes at matched points
The steeper the original, the flatter the inverse, and a horizontal tangent on one reflects to a vertical tangent on the other — which is exactly why the theorem needs the original's derivative to be non-zero.
Worked example
Example 3.52. The argument of f prime is the whole difficulty.
\[ \text{Let } f(x) = x^{3}+2x. \text{ Given } f(1) = 3, \text{ find } (f^{-1})'(3). \]
Write the theorem
Why: The reciprocal, at the inverse's output.
Find the inverse's output at 3
Why: Since f of 1 is 3, the inverse of 3 is 1.
Differentiate the original
Why: Termwise.
\[ f'(x) = 3 x ^{2} + 2 \]
Evaluate at that output
Why: At 1, not at 3.
\[ f'(1) = 5 \]
Take the reciprocal
Why: The theorem.
\[ \frac{1}{5} \]
Figure (svg): The solution to Worked example an inverse's derivative at a point shown as a ladder of expressions, one row per legal move
\[ \big(f^{-1}\big)'(3) = \tfrac{1}{5} \]
Verify: check what evaluating at the wrong input would give
Why: Evaluating f prime at 3 instead of 1 gives 29, and the reciprocal one twenty-ninth — a completely different answer. The correct input is the point ON THE ORIGINAL CURVE that corresponds to 3, which is 1. Note also that the inverse of this cubic cannot be written down in any convenient form, and the theorem never needed it: only the single value f inverse of 3 was required, and that came from reading the given fact backwards.
OpenStax Calculus Volume 1, §3.7 Derivatives of Inverse Functions §3.7, pp. 261-262
Fill the middle
The cubic from the worked example, with f of 1 equal to 3.
Fill in the blanks
(f^1)'(3) = \frac______(3)\big)} = \frac______})}
Why: Since f of 1 is 3, the inverse of 3 is 1, so f prime is evaluated at 1. Evaluating at 3 instead would answer a different question entirely.
Worked example
Checkpoint 3.52. A case where both derivatives are computable.
\[ \text{For } f(x) = x^{2} \text{ on } [0,\infty) \text{ with inverse } \sqrt{x}, \text{ verify the theorem at } x = 9. \]
Compute the inverse's derivative directly
Why: The root's derivative from Section 3.3.
\[ \frac{1}{2 \sqrt{9}} = \frac{1}{6} \]
Now use the theorem instead
Why: First find the inverse's output.
Differentiate the original
Why: Power rule.
\[ f'(x) = 2 x \]
Evaluate at that output and reciprocate
Why: At 3, giving 6.
\[ \frac{1}{6} \]
Compare
Why: Identical.
Figure (svg): The solution to Worked example checking with a known pair shown as a ladder of expressions, one row per legal move
\[ \big(f^{-1}\big)'(9) = \frac{1}{f'(3)} = \frac{1}{6} \]
Verify: see the reciprocal-slope picture in the numbers
Why: The squaring function has slope 6 at the point (3, 9), and the square root has slope one sixth at the reflected point (9, 3). The two tangent lines are reflections of each other in the diagonal, and their slopes are reciprocals — which is exactly what the picture predicted. Having a case where both sides are independently computable is what makes the theorem believable before it is used on inverses that cannot be written down.
OpenStax Calculus Volume 1, §3.7 Derivatives of Inverse Functions §3.7, pp. 262-262
Trap
\[ f(x) = x^{3}+2x, \quad (f^{-1})'(3) = \frac{1}{f'(3)} \]
Substitute 3 into f prime
Why: The student matches the number in the question.
\[ = \frac{1}{29} \quad \text{(wrong)} \]
The input 3 belongs to the inverse. The original's derivative must be evaluated at the corresponding point on the original's own graph, which is 1.
\[ (f^{-1})'(3) = \frac{1}{f'\big(f^{-1}(3)\big)} = \frac{1}{f'(1)} = \tfrac{1}{5} \]
Find the inverse's output first, then evaluate f prime there
Why: The two functions live on reflected points, so they take different inputs.
The picture keeps it straight: the point on the inverse is (3, 1) and the corresponding point on the original is (1, 3). The original's tangent is drawn at input 1, so its slope is f prime of 1. Writing both points down before computing anything makes the error nearly impossible.
Matching
Reflection swaps the coordinates.
Match the pairs
Why: The last row is why the theorem needs a non-zero derivative: a horizontal tangent reciprocates into a vertical one, and a vertical tangent means no derivative at all. The other rows are the coordinate swap and its consequence for slopes.
Two truths and a lie
All three are about the theorem.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it is the section's characteristic error. The original's derivative must be evaluated at the INVERSE'S OUTPUT, not at x. For the cubic example that is 1 rather than 3, and the two give one fifth and one twenty-ninth — completely different numbers.
Prediction
Commit before reasoning.
Predict first
What goes wrong if f has a horizontal tangent at the corresponding point?
Correct: The reciprocal is undefined, and the inverse has a vertical tangent.
\[ f(x) = x^3: \; f'(0) = 0 \;\Longrightarrow\; \text{the cube root has a vertical tangent at } 0 \]
Why: Reflection turns a horizontal tangent into a vertical one, and a vertical tangent means the difference quotient is unbounded — no derivative. The formula's division by zero is the algebra recording that geometry. The inverse itself may exist perfectly well: the cubing function has a horizontal tangent at the origin and its inverse, the cube root, is defined everywhere and merely has a vertical tangent at 0. Existence of the inverse and differentiability of it are separate questions.
Section
Section 2
Concept
The nth root is the inverse of the nth power, so the theorem gives its derivative. Combining that with the chain rule extends the power rule to every rational exponent, completing a claim Section 3.3 made but did not earn.
the rational power rule — For any rational exponent p over q, the derivative of x to that power is the exponent times x to the exponent minus one — the same formula as for whole exponents, now proved.
\[ \frac{d}{dx}\left[x^{p/q}\right] = \frac{p}{q}x^{p/q - 1} \]
The derivation is worth seeing because it is the pattern for Section 3.8's implicit differentiation: raise both sides to remove the root, differentiate, and solve for the derivative.
Figure (svg): The power rule extended to a rational exponent, derived from the inverse of a power
OpenStax Calculus Volume 1, §3.7 Derivatives of Inverse Functions §3.7, pp. 262-264 — extending the power rule
Picture it
Five lines from the inverse relationship to the general formula.
Figure (svg): The power rule extended to a rational exponent, derived from the inverse of a power
The third line is the chain rule doing the work: differentiating y to the n with respect to x produces the factor y prime, which is exactly what is being solved for.
Worked example
Example 3.54. Raise, differentiate, solve.
\[ \text{Find } \frac{d}{dx}\left[x^{1/3}\right] \text{ from the inverse relationship.} \]
Name the function and remove the root
Why: Cube both sides.
\[ y = x ^{\frac{1}{3}},\text{ so } y ^{3} = x \]
Differentiate both sides with respect to x
Why: The chain rule on the left.
\[ 3 y ^{2} y' = 1 \]
Solve for the derivative
Why: Divide.
\[ y' = \frac{1}{3 y ^{2}} \]
Substitute back for y
Why: y is the cube root of x.
\[ \frac{1}{3 x ^{\frac{2}{3}}} \]
Rewrite as a power
Why: Matching the power rule's form.
\[ (\frac{1}{3}) x ^{-\frac{2}{3}} \]
Figure (svg): The solution to Worked example deriving the root's derivative shown as a ladder of expressions, one row per legal move
\[ \frac{d}{dx}\left[x^{1/3}\right] = \tfrac{1}{3}x^{-2/3} \]
Verify: check against the power rule's prediction and the graph
Why: The power rule with exponent one third predicts one third times x to the negative two thirds — matching exactly, which is what the derivation was meant to establish rather than assume. The answer is positive for every non-zero x, matching a cube root that increases everywhere; and it is unbounded as x approaches 0, matching the vertical tangent there. All three checks agree.
OpenStax Calculus Volume 1, §3.7 Derivatives of Inverse Functions §3.7, pp. 263-263
Fill the middle
Differentiating a two-thirds power.
Fill in the blanks
\frac-1/3___\left[x^___\right] = \tfrac______x^___}
Why: Two thirds minus one is negative one third. The negative exponent means the derivative is unbounded near 0, which matches the cusp the graph has there.
Worked example
Checkpoint 3.54. Combine with Section 3.6.
\[ \text{Differentiate } y = (x^{2}+1)^{2/3}. \]
Apply the power rule with a rational exponent
Why: Exponent down, reduced by one.
\[ (\frac{2}{3}) (x ^{2} + 1) ^{-\frac{1}{3}} \]
Multiply by the inner derivative
Why: The chain rule.
\[ \times 2 x \]
Combine
Why: Collect the constants.
\[ (4 x / 3) (x ^{2} + 1) ^{-\frac{1}{3}} \]
Rewrite without a negative exponent
Why: As a fraction.
\[ 4 x / (3(x ^{2} + 1) ^{\frac{1}{3}}) \]
Figure (svg): The solution to Worked example a rational power with the chain rule shown as a ladder of expressions, one row per legal move
\[ y' = \frac{4x}{3\,(x^{2}+1)^{1/3}} \]
Verify: check the sign and a value
Why: The derivative is negative for negative x and positive for positive x, matching a curve with a minimum at the origin — which the function has, since x squared plus 1 is smallest there. At x equal to 0 the derivative is 0, confirming the horizontal tangent. Reducing the exponent two thirds by one gives negative one third, and getting that subtraction wrong is the standard slip with rational exponents.
OpenStax Calculus Volume 1, §3.7 Derivatives of Inverse Functions §3.7, pp. 264-264
Error analysis
A student differentiates a two-thirds power.
Annotate
On: \( \frac{d}{dx}\left[x^{2/3}\right] = \tfrac{2}{3}x^{1/3} \)
The graph settles it: x to the two thirds has a cusp at the origin with unbounded slope, so its derivative cannot tend to zero there. Writing the subtraction as a fraction over a common denominator prevents the slip.
Ranking
Getting the root's derivative from the inverse relationship.
Put in order
Why: Step c is where the chain rule is essential: differentiating y to the n with respect to x produces n y to the n minus 1 times y prime, and that y prime is what step d isolates. This is exactly the method Section 3.8 will generalise as implicit differentiation.
Matching
Rational exponents, by the same rule.
Match the pairs
Why: Every one is the same rule with the subtraction done carefully. The last shows a negative exponent becoming more negative, which is where sign errors accumulate — writing the subtraction over a common denominator each time prevents them.
Prediction
Commit before reasoning.
Predict first
Section 3.3 already stated the power rule for all real exponents. What has this section added?
Correct: A proof for the rational case, which was previously only asserted.
\[ \text{whole } n: \text{ binomial}; \quad \text{rational } p/q: \text{ this section}; \quad \text{real: Section 3.9} \]
Why: Section 3.3 proved the rule for whole-number exponents from the binomial expansion and stated it for all real ones. The rational case genuinely needed this argument, because a root is an inverse and its derivative requires the inverse function theorem. The formula is unchanged, which is the point — the same rule, now earned rather than assumed. The irrational case waits for Section 3.9's logarithmic differentiation.
Section
Section 3
Concept
Differentiating the cancellation equation for sine and arcsine gives the derivative in terms of a cosine of an arcsine. Drawing a right triangle converts that composition into an algebraic expression, and the arcsine disappears entirely.
the triangle argument — Setting theta equal to the inverse trigonometric function makes the defining ratio one of the triangle's sides over another. Pythagoras supplies the third side, and every other trigonometric function of theta can then be read off as algebra.
\[ \frac{d}{dx}\left[\arcsin x\right] = \frac{1}{\sqrt{1-x^{2}}} \]
The result is striking. The arcsine is a transcendental function and its derivative is algebraic — no trigonometry survives. The same happens for all six, which is why they turn up as antiderivatives in Section 5.7.
Figure (svg): The right triangle that converts a trigonometric function of an inverse trigonometric function into algebra
OpenStax Calculus Volume 1, §3.7 Derivatives of Inverse Functions §3.7, pp. 264-266 — derivatives of inverse trigonometric functions
Picture it
Theta is the angle whose sine is x, drawn.
Figure (svg): The right triangle that converts a trigonometric function of an inverse trigonometric function into algebra
With the hypotenuse taken as 1 and the opposite side as x, the adjacent side is the root of 1 minus x squared by Pythagoras — and that is the cosine of theta, which is exactly what the derivative needed.
Worked example
Example 3.56. Differentiate the cancellation equation.
\[ \text{Find } \frac{d}{dx}\left[\arcsin x\right]. \]
Write the cancellation equation
Why: Sine undoes arcsine.
\[ \sin(\arcsin x) = x \]
Differentiate both sides with the chain rule
Why: Cosine of the inside, times the inside's derivative.
\[ \cos(\arcsin x) \times(\arcsin)' = 1 \]
Solve for the derivative
Why: Divide.
\[ (\arcsin)' = 1 / \cos(\arcsin x) \]
Draw the triangle to simplify the cosine
Why: Opposite x, hypotenuse 1, adjacent the root of 1 minus x squared.
\[ \cos(\arcsin x) = \sqrt{1 - x ^{2}} \]
Substitute
Why: The arcsine has vanished.
\[ 1 / \sqrt{1 - x ^{2}} \]
Figure (svg): The solution to Worked example deriving the arcsine derivative shown as a ladder of expressions, one row per legal move
\[ \frac{d}{dx}\left[\arcsin x\right] = \frac{1}{\sqrt{1-x^{2}}} \]
Verify: check the sign, the domain, and one value
Why: The expression is positive throughout, matching an arcsine that increases across its whole domain. At x equal to 0 it gives 1, and the arcsine does pass through the origin at 45 degrees. As x approaches plus or minus 1 the denominator vanishes and the derivative is unbounded — matching the vertical tangents at the endpoints, which are the reflections of sine's horizontal tangents at plus and minus pi over 2. The positive square root was chosen because the arcsine's range lies where cosine is non-negative.
OpenStax Calculus Volume 1, §3.7 Derivatives of Inverse Functions §3.7, pp. 265-265
Fill the middle
The triangle for theta equal to the arcsine of x, with hypotenuse 1.
Fill in the blanks
\sin\theta = x, \; \text\sqrt{1-x^2} 1 \;\Longrightarrow\; \cos\theta = ___
Why: With opposite x and hypotenuse 1, the adjacent side is the root of 1 minus x squared, which is the cosine. That substitution is what removes the arcsine from the answer.
Worked example
Checkpoint 3.56. Same method, different triangle.
\[ \text{Find } \frac{d}{dx}\left[\arctan x\right]. \]
Write the cancellation equation
Why: Tangent undoes arctangent.
\[ \tan(\arctan x) = x \]
Differentiate with the chain rule
Why: The derivative of tangent is secant squared.
\[ \sec ^{2}(\arctan x) \times(\arctan)' = 1 \]
Solve
Why: Divide.
\[ 1 / \sec ^{2}(\arctan x) \]
Use the identity rather than a triangle
Why: Secant squared is 1 plus tangent squared.
\[ \sec ^{2} = 1 + x ^{2} \]
Substitute
Why: The arctangent has vanished.
\[ \frac{1}{1 + x ^{2}} \]
Figure (svg): The arctangent with its derivative beneath, showing the bounded function with a positive shrinking slope
\[ \frac{d}{dx}\left[\arctan x\right] = \frac{1}{1+x^{2}} \]
Verify: check against the graph's shape
Why: The derivative is positive everywhere, so the arctangent increases everywhere — which it does. It is largest at x equal to 0, where it equals 1, and the arctangent is steepest there. And it tends to 0 as x grows, which is exactly what a function levelling off toward a horizontal asymptote must do. The identity shortcut avoided drawing a triangle, and it works here because secant squared relates directly to tangent squared, which the equation already supplies.
OpenStax Calculus Volume 1, §3.7 Derivatives of Inverse Functions §3.7, pp. 266-266
Trap
\[ \frac{d}{dx}\left[\arcsin x\right] = \frac{1}{\cos(\arcsin x)} \]
Stop after solving for the derivative
Why: The student treats this as a finished answer.
It is correct but useless: evaluating it requires computing an arcsine and then a cosine, and it hides the fact that the answer is algebraic.
\[ \frac{1}{\cos(\arcsin x)} = \frac{1}{\sqrt{1-x^{2}}} \]
Draw the triangle and eliminate the inverse function
Why: The composition of a trigonometric function with an inverse trigonometric one is always algebraic.
Simplifying is not cosmetic here. It reveals that the derivative of a transcendental function is algebraic, which is why these expressions appear as antiderivatives in Section 5.7 — a fact entirely hidden by the unsimplified form.
Ranking
Finding an inverse trigonometric derivative.
Put in order
Why: Step d is the one that turns a correct answer into a useful one, and it is where the result's algebraic character becomes visible. Step e catches the sign choice, which matters because the triangle argument requires picking a square root.
Two truths and a lie
All three are about these derivatives.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it drops an entire layer. The correct intermediate form is one over the cosine of the ARCSINE of x, not of x itself — the chain rule evaluates the outer derivative at the inner output. Simplifying that gives one over the root of 1 minus x squared, which is a very different function from the secant.
Prediction
Commit before reasoning.
Predict first
Why does no trigonometry survive in the derivative of arcsine?
Correct: Because a trigonometric function of an inverse trigonometric function is always algebraic.
\[ \cos(\arcsin x) = \sqrt{1-x^2}, \quad \tan(\arcsin x) = \frac{x}{\sqrt{1-x^2}} \]
Why: Setting theta equal to the arcsine makes sine of theta equal to x, and then Pythagoras gives every other ratio in the triangle as an algebraic expression in x. The arcsine itself is emphatically transcendental — no finite algebraic formula produces it. The chain rule creates the composition rather than removing it; the triangle is what evaluates it. This is why Section 5.7 finds these algebraic expressions turning up as integrands whose antiderivatives are inverse trigonometric.
Section
Section 4
Concept
The same argument gives all six inverse trigonometric derivatives, and they fall into three pairs. Each co-function's derivative is the negative of its partner's, because the two functions sum to a constant.
the complementary identity — The arcsine and arccosine sum to pi over two, so differentiating gives derivatives that are negatives of each other. The same relationship holds for the other two pairs.
\[ \arcsin x + \arccos x = \tfrac{\pi}{2} \;\Longrightarrow\; (\arccos)' = -(\arcsin)' \]
That identity is a genuine shortcut. Once the arcsine's derivative is known, the arccosine's follows in one line without any triangle at all, and the same works for the other two pairs.
Figure (svg): The derivatives of the six inverse trigonometric functions, grouped by the co-pattern
OpenStax Calculus Volume 1, §3.7 Derivatives of Inverse Functions §3.7, pp. 265-267 — the six inverse trigonometric derivatives
Picture it
The complete table.
Figure (svg): The derivatives of the six inverse trigonometric functions, grouped by the co-pattern
Three distinct algebraic forms, each appearing twice with opposite signs. The absolute value in the last pair is genuine and is there because the arcsecant's domain has two separated pieces.
Worked example
Example 3.58. Use the complementary identity.
\[ \text{Find } \frac{d}{dx}\left[\arccos x\right] \text{ without a triangle.} \]
Write the complementary identity
Why: The two angles sum to a right angle.
\[ \arcsin x + \arccos x = \frac{\pi}{2} \]
Differentiate both sides
Why: The right side is a constant.
\[ (\arcsin)' + (\arccos)' = 0 \]
Solve
Why: The arccosine's derivative is the negative.
\[ (\arccos)' = -(\arcsin)' \]
Substitute the known derivative
Why: From the previous idea.
\[ -1 / \sqrt{1 - x ^{2}} \]
Figure (svg): The solution to Worked example the arccosine in one line shown as a ladder of expressions, one row per legal move
\[ \frac{d}{dx}\left[\arccos x\right] = -\frac{1}{\sqrt{1-x^{2}}} \]
Verify: check the sign against the graph
Why: The arccosine decreases from pi at x equal to negative 1 to 0 at x equal to 1, so its derivative must be negative throughout — and it is. The identity shortcut cost one line where the triangle argument would have cost five, and it also explains WHY the two derivatives differ only in sign, which the triangle argument would have produced as an unexplained coincidence.
OpenStax Calculus Volume 1, §3.7 Derivatives of Inverse Functions §3.7, pp. 266-267
Matching
Three forms, each appearing twice.
Match the pairs
Why: Each pair differs only in sign, because the two functions in a pair sum to a constant. That relationship halves what has to be learned and explains the minus signs rather than leaving them to be memorised.
Worked example
Checkpoint 3.58. The inner function is not a bare x.
\[ \text{Differentiate } y = \arctan(3x^{2}). \]
Apply the arctangent derivative to the outer layer
Why: One over 1 plus the inside squared.
\[ \frac{1}{1 + (3 x ^{2}) ^{2}} \]
Simplify the inside
Why: Nine x to the fourth.
\[ \frac{1}{1 + 9 x ^{4}} \]
Multiply by the inner derivative
Why: The chain rule.
\[ \times 6 x \]
Combine
Why: Collect.
\[ 6 x / (1 + 9 x ^{4}) \]
Figure (svg): The solution to Worked example an inverse trigonometric with the chain rule shown as a ladder of expressions, one row per legal move
\[ y' = \frac{6x}{1+9x^{4}} \]
Verify: check the sign and the value at the origin
Why: The derivative is negative for negative x and positive for positive x, matching a function with a minimum at the origin — which arctangent of 3x squared has, since its input is smallest there. At x equal to 0 the derivative is 0, confirming the horizontal tangent. Note the inside was SQUARED in the denominator, not left as 3x squared: the formula calls for 1 plus the argument squared, and forgetting to square is the standard slip here.
OpenStax Calculus Volume 1, §3.7 Derivatives of Inverse Functions §3.7, pp. 267-267
Error analysis
A student differentiates an arctangent of a composite argument.
Annotate
On: \( \frac{d}{dx}\left[\arctan(3x^{2})\right] = \frac{6x}{1 + 3x^{2}} \)
The denominator of an arctangent derivative always contains the square of whatever the arctangent received. A reduction check catches it: with a bare x inside, the denominator must be 1 plus x squared, and the pattern must extend consistently.
Fill the middle
Differentiating the arctangent of a composite argument.
Fill in the blanks
\frac9x^4___\left[\arctan(3x^___)\right] = \frac______}}
Why: The denominator is 1 plus the square of the argument, and the square of 3x squared is 9x to the fourth. Leaving it as 3x squared is the standard error.
Sorting
Apply the co-pattern.
Sort into buckets
Sort each inverse trigonometric function.
The pattern matches Section 3.5's exactly: the three names carrying co- are the three derivatives with a minus. Here the reason is sharper, since each pair provably sums to a constant.
Prediction
Commit before reasoning.
Predict first
Why is the arccosine's derivative exactly the negative of the arcsine's?
Correct: Because they sum to the constant pi over two, so their derivatives sum to zero.
\[ \arcsin x + \arccos x = \tfrac{\pi}{2} \;\Longrightarrow\; (\arcsin)' + (\arccos)' = 0 \]
Why: Differentiating a constant gives zero, so the two derivatives must be negatives. This is a genuine explanation rather than a coincidence, and it turns the second derivative into a one-line consequence of the first. Cosine is certainly not the negative of sine, and sharing a domain implies nothing about derivatives — two functions can share a domain and behave entirely differently.
Section
Section 5
Concept
The arcsine and arccosine are defined on the closed interval from negative one to one but differentiable only on its interior, because their graphs have vertical tangents at the endpoints. The arctangent, by contrast, is differentiable everywhere.
endpoint behaviour — At an endpoint of an inverse trigonometric function's domain, the original function had a horizontal tangent, so the reflection has a vertical one and no derivative exists there.
\[ \arcsin: \text{ defined on } [-1,1], \text{ differentiable on } (-1,1) \]
This is the inverse function theorem's non-vanishing condition made concrete. Sine has horizontal tangents at plus and minus pi over two, and those reflect into the arcsine's vertical tangents at plus and minus one.
Figure (svg): The domains of the inverse trigonometric derivatives, showing where each fails
OpenStax Calculus Volume 1, §3.7 Derivatives of Inverse Functions §3.7, pp. 265-267 — domains of the derivatives
Picture it
Where each of four inverse functions is differentiable.
Figure (svg): The domains of the inverse trigonometric derivatives, showing where each fails
In every bounded case the derivative's domain is the function's with the endpoints removed. The arctangent escapes because it never reaches its horizontal asymptotes, so there is no endpoint to lose.
Worked example
Example 3.59. The formula and the picture agree.
\[ \text{Explain why } \arcsin \text{ is not differentiable at } x = 1. \]
Evaluate the derivative formula near the endpoint
Why: The denominator involves 1 minus x squared.
\[ 1 / \sqrt{1 - x ^{2}} \]
See what happens as x approaches 1
Why: The radicand approaches 0.
Conclude for the formula
Why: Division by zero.
Explain it geometrically
Why: Sine has a horizontal tangent at pi over 2.
State the domains
Why: Defined at 1, not differentiable there.
\[ [-1, 1]\text{ versus } (-1, 1) \]
Figure (svg): The solution to Worked example the endpoint behaviour of arcsine shown as a ladder of expressions, one row per legal move
\[ \lim_{x \to 1^-}\frac{1}{\sqrt{1-x^{2}}} = \infty \]
Verify: confirm the reflection explanation numerically
Why: Sine's derivative at pi over 2 is cosine of pi over 2, which is 0. The inverse function theorem's reciprocal is therefore undefined there, exactly as the formula shows. The arcsine is still perfectly well defined at 1, with value pi over 2 — being defined and being differentiable are separate questions, as Section 3.2 established. This is the theorem's non-vanishing condition failing in the simplest possible case.
OpenStax Calculus Volume 1, §3.7 Derivatives of Inverse Functions §3.7, pp. 266-266
Sorting
Check whether the derivative's denominator vanishes.
Sort into buckets
Sort each case.
The arctangent is differentiable at every real number, however large, because its denominator is at least 1. The bounded inverses each lose exactly their two endpoints, and those are the points where their partners had horizontal tangents.
Worked example
Checkpoint 3.59. No endpoints, no failures.
\[ \text{Explain why } \arctan \text{ is differentiable on all of } \mathbb{R}. \]
Examine the derivative's denominator
Why: One plus x squared.
\[ 1 + x ^{2} \]
Ask when it vanishes
Why: A square is never negative.
Conclude for the formula
Why: No division by zero anywhere.
Explain geometrically
Why: Tangent's derivative, secant squared, is never zero.
Figure (svg): The solution to Worked example why the arctangent is different shown as a ladder of expressions, one row per legal move
\[ 1 + x^{2} \ge 1 > 0 \text{ for every real } x \]
Verify: trace it back to the theorem's condition
Why: The inverse function theorem needs tangent's derivative to be non-zero, and secant squared is at least 1 wherever it is defined — never zero. So no point reflects into a vertical tangent, and the arctangent is smooth throughout. Contrast the arcsine, whose partner sine has two horizontal tangents per period, each producing an endpoint failure. The behaviour of the derivative is entirely determined by where the ORIGINAL function flattens.
OpenStax Calculus Volume 1, §3.7 Derivatives of Inverse Functions §3.7, pp. 267-267
Trap
\[ \arcsin \text{ is defined on } [-1, 1] \]
Conclude its derivative is defined there too
Why: The student carries the domain across.
\[ \left.\frac{1}{\sqrt{1-x^2}}\right|_{x=1} = \frac{1}{0} \quad \text{(undefined)} \]
The graph has a vertical tangent at the endpoint, so the difference quotient is unbounded and no derivative exists.
\[ \text{defined on } [-1,1], \text{ differentiable on } (-1,1) \]
Determine the derivative's domain from the derivative itself
Why: It is where the formula is defined, which may be a strictly smaller set.
This is the same lesson as Section 3.2's square root: a function can be defined and continuous at a point where its derivative is not. It matters practically in Chapter 4, where extreme values are hunted at exactly the points where a derivative fails to exist as well as where it vanishes.
Fill the middle
The arcsine's derivative, and where its denominator vanishes.
Fill in the blanks
\frac\pm 1___}} \text___ x = ___
Why: At plus and minus 1 the radicand vanishes and the derivative is unbounded. Those are exactly the endpoints of the arcsine's domain, where its graph has vertical tangents.
Two truths and a lie
All three are about domains.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it is the same error as assuming continuity gives differentiability. The arcsine at 1, the absolute value at 0 and the cube root at 0 are all defined and all fail to have derivatives. A derivative's domain must always be determined from the derivative itself.
Prediction
Commit before reasoning.
Predict first
Why does the arcsine have a vertical tangent at x = 1?
Correct: Because sine has a horizontal tangent there and reflection turns horizontal into vertical.
\[ \sin'\!\left(\tfrac{\pi}{2}\right) = 0 \;\Longrightarrow\; \text{reciprocal undefined} \;\Longrightarrow\; \text{vertical tangent} \]
Why: The point (pi over 2, 1) on sine reflects to (1, pi over 2) on arcsine, and sine's tangent there is horizontal because cosine of pi over 2 is zero. Reflecting a horizontal line in the diagonal gives a vertical one. That is the geometry, and the formula's vanishing denominator is the algebra recording it — the inverse function theorem's non-zero condition failing exactly where it must. The domain ending at 1 is a consequence of the same fact, not the cause.
Comparison
Fill the blanks. Each pair shares an algebraic form and differs only in sign.
Comparison matrix
| Pair | Shared form | Why they differ only in sign |
|---|---|---|
| arcsin, arccos | 1/sqrt(1-x^2) | they sum to pi/2, a constant |
| arctan, arccot | 1/(1+x^2) | they sum to pi/2, a constant |
| arcsec, arccsc | 1/(|x| sqrt(x^2-1)) | they sum to pi/2, a constant |
| All six | algebraic | a trig function of an inverse trig function simplifies by a triangle |
The last row is the section's real surprise. Six transcendental functions, and every one of their derivatives is an ordinary algebraic expression — which is why Section 5.7 finds them appearing as antiderivatives.
Pattern
Given an inverse function to differentiate.
For a numerical question about an inverse's derivative at a point, use the theorem directly — but find the inverse's OUTPUT first, and evaluate the original's derivative there rather than at the given input.
Stewart, Calculus: Early Transcendentals 8e, §3.5 Implicit Differentiation §3.5, pp. 208-217
Check
The theorem. Evaluate at the right input.
Check your understanding
For f(x) = x^3 + 2x with f(1) = 3, find (f inverse)'(3).
Answer: A
Why: The inverse of 3 is 1, and f'(1) = 5, so the reciprocal is 1/5.
Check
The arcsine. The result is algebraic.
Check your understanding
What is the derivative of arcsin x?
Answer: A
Why: Differentiating the cancellation equation gives 1 over cosine of the arcsine, which the triangle turns into this.
Check
Domains. Determine them from the derivative.
Check your understanding
On what set is arcsin x differentiable?
Answer: A
Why: At the endpoints the denominator vanishes and the graph has vertical tangents.
Real world
A camera on the ground films a rocket rising vertically from a launch pad 500 metres away. The camera's angle of elevation must track the rocket, and the operator needs to know how fast to rotate it.
Discussion prompt
Express the elevation angle as an inverse trigonometric function of the rocket's height, differentiate it, and say what happens to the required rotation rate as the rocket climbs very high.
Hint: The tangent of the angle is the height over the horizontal distance.
Answer:
\[ \tan\theta = \frac{h}{500} \;\Longrightarrow\; \theta = \arctan\!\left(\frac{h}{500}\right) \]
Differentiating with respect to the height, using the arctangent derivative and the chain rule:
\[ \frac{d\theta}{dh} = \frac{1}{1 + (h/500)^{2}}\cdot\frac{1}{500} = \frac{500}{250000 + h^{2}} \]
At launch, with the height near zero, this is one over 500 radians per metre — the angle changes fastest at the start.
As the rocket climbs, the denominator grows and the rate falls toward zero. At a height of 5000 metres it is about a hundredth of its initial value. The camera must swing quickly at first and barely move once the rocket is high, which is exactly what the arctangent's flattening graph predicts.
Note the chain rule supplied the factor of one over 500, and the arctangent derivative supplied the rest — and the answer is entirely algebraic, with no trigonometry left in it. That is what makes it usable by a control system, which is a large part of why these derivatives matter.
Commit first
Answer, then rate your confidence honestly.
Predict first
For f with f(1) = 3, what is (f inverse)'(3)?
Correct: One over f prime of 1.
\[ f(1) = 3 \;\Longrightarrow\; f^{-1}(3) = 1 \;\Longrightarrow\; (f^{-1})'(3) = \frac{1}{f'(1)} \]
Why: The theorem evaluates the original's derivative at the INVERSE'S OUTPUT, and since f of 1 is 3, that output is 1. Evaluating at 3 is the section's characteristic error and gives a completely different number — for the cubic example, one twenty-ninth instead of one fifth. The reciprocal is also essential: forgetting it inverts the relationship, and the picture explains why it must be there, since reflection turns a slope into its reciprocal.
Explain it
They keep evaluating f prime at the number in the question rather than at the inverse's output.
Discussion prompt
In four sentences or fewer, give them a picture that fixes it.
Hint: Have them plot both points.
Answer:
Have them mark the point (1, 3) on the original curve and its reflection (3, 1) on the inverse. The tangent they want is at the point on the INVERSE, which sits above the input 3 — and its reflection is the tangent on the original, which sits above the input 1.
So the original's derivative is taken at 1, because that is where its own tangent lives. The number 3 is an input for the inverse and an OUTPUT for the original, and putting it into f prime asks about a completely different point on the curve.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the input, write both reflected points down before computing anything. For the triangle, set theta equal to the inverse function, label the two sides the definition gives, and get the third from Pythagoras. For signs, remember that co-functions decrease, so their derivatives are negative. For domains, read them off the derivative's own formula rather than the function's. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, draw a function and its inverse reflected in the dashed diagonal, with tangents at a matched pair of points, and label the two slopes as reciprocals. Beside it write the theorem, circling the argument of f prime and writing one sentence on why it is not x. Below, take the cubic x cubed plus 2x with f of 1 equal to 3 and compute the inverse's derivative at 3 in full, writing both reflected points first. In the middle of the page, derive the arcsine's derivative completely: the cancellation equation, the chain rule step, solving, and then the right triangle with all three sides labelled. Then get the arccosine's derivative from the complementary identity in one line. At the bottom, write all six inverse trigonometric derivatives in a table, circling the three with minus signs, and beside the table write the domain of each derivative alongside the domain of its function. In a margin, draw sine's horizontal tangent at pi over two and its reflection, and write why the arcsine loses its endpoints.
If your six-derivative table has any trigonometric function in it, something has been left unsimplified — every one of these derivatives is algebraic, and that is the section's most surprising result.
Recap
Five things, and all of them come from the chain rule applied to one equation.
| If you see | Then |
|---|---|
| An inverse's derivative at a point | Find the inverse's output first |
| A horizontal tangent on f | A vertical tangent on the inverse: no derivative |
| A root to differentiate | It is an inverse; the rational power rule applies |
| A trig function of an inverse trig function | Draw a triangle; the result is algebraic |
| A co- inverse trig function | Its derivative carries a minus sign |
| arctan of something | Square that something in the denominator |
| An endpoint of arcsin or arccos | Defined but not differentiable |
Section 3.8 generalises the technique used here for the root: differentiate an equation as it stands, treating y as a function of x, and solve for the derivative. That is implicit differentiation, and it handles curves that are not functions at all.
OpenStax Calculus Volume 1, §3.7 Derivatives of Inverse Functions §3.7, pp. 259-267 — everything on these slides traces back here
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