Differentiating an equation in x and y without solving for y, the chain-rule factor every y term produces, tangent lines to curves that fail the vertical line test, reading horizontal and vertical tangents off an implicit derivative, and computing a second derivative implicitly.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 3 — Derivatives
Implicit Differentiation
Objectives
Five outcomes. The technique is the same one Section 3.7 used on a root, now stated generally and applied to curves no function can describe.
OpenStax Calculus Volume 1, §3.8 Implicit Differentiation §3.8, pp. 268-274 — the section these objectives are drawn from
Warm-up
Section 3.7 differentiated the cube root by writing y equal to the root, cubing both sides, and differentiating the equation that resulted.
Discussion prompt
That derivation never solved for anything — it differentiated an equation containing y and then solved for y prime. Why did differentiating y cubed produce a factor of y prime?
Hint: Ask what y is a function of.
Answer:
\[ y^{3} = x \;\Longrightarrow\; 3y^{2}\frac{dy}{dx} = 1 \]
Because y is a function of x, differentiating y cubed with respect to x is a composition: the outer cube applied to the inner function y. The chain rule then supplies the factor dy by dx.
That single observation is the whole of implicit differentiation. It works on any equation relating x and y, including ones that cannot be solved for y at all — which is what this section is really for.
Concept
An equation in x and y can be differentiated term by term without first solving for y. Every y term produces a chain-rule factor dy by dx, and the resulting equation is then solved for that factor.
implicit differentiation — Differentiating both sides of an equation in x and y with respect to x, treating y as an unspecified function of x, and then solving the result for dy by dx.
\[ \frac{d}{dx}\left[y^{n}\right] = ny^{n-1}\frac{dy}{dx} \]
The name is slightly misleading: nothing is done implicitly. The equation is differentiated perfectly explicitly, and it is the FUNCTION that is left implicit rather than the differentiation.
Figure (svg): The chain-rule factor that every y term produces, shown side by side with an x term
OpenStax Calculus Volume 1, §3.8 Implicit Differentiation §3.8, pp. 268-270
Section
Section 1
Concept
Differentiating an x term is ordinary. Differentiating a y term is a composition, because y is a function of x, so the chain rule attaches a factor dy by dx. Omitting that factor destroys the method entirely.
the implicit factor — The dy by dx produced whenever a term containing y is differentiated with respect to x. It arises from the chain rule, since y is itself a function of x.
\[ \frac{d}{dx}\left[x^{2}\right] = 2x \quad \text{but} \quad \frac{d}{dx}\left[y^{2}\right] = 2y\frac{dy}{dx} \]
A useful mental substitution is to write y as f of x while differentiating. Then y squared is visibly the composition f of x, squared, and the chain rule's factor is obviously required.
Figure (svg): The chain-rule factor that every y term produces, shown side by side with an x term
OpenStax Calculus Volume 1, §3.8 Implicit Differentiation §3.8, pp. 268-271 — implicit differentiation
Picture it
An x term and a y term differentiated side by side.
Figure (svg): The chain-rule factor that every y term produces, shown side by side with an x term
The only difference is the trailing factor, and it is the difference between a method that works and one that produces nonsense. Every term must be checked for a y before it is differentiated.
Worked example
Example 3.60. Four steps, and the y terms are the ones to watch.
\[ \text{Find } \frac{dy}{dx} \text{ for } x^{2}+y^{2}=25. \]
Differentiate both sides with respect to x
Why: Term by term.
\[ d / \,dx [x ^{2}] + d / \,dx [y ^{2}] = d / \,dx [25] \]
Handle the x term normally
Why: Power rule.
\[ 2 x \]
Handle the y term with the chain rule
Why: The extra factor appears.
\[ 2 y(\,dy / \,dx) \]
Collect and solve
Why: The right side is zero.
\[ 2 y(\,dy / \,dx) = -2 x \]
Divide
Why: The derivative.
\[ \,dy / \,dx = -\frac{x}{y} \]
Figure (svg): The four steps of implicit differentiation, applied to a circle
\[ \frac{dy}{dx} = -\frac{x}{y} \]
Verify: check against the explicit route on the upper half
Why: Solving for the upper semicircle gives y equal to the root of 25 minus x squared, whose derivative by the chain rule is minus x over that root — which is minus x over y exactly. The implicit answer covers BOTH halves at once, whereas the explicit route needed the branch chosen first. Geometrically the answer is right too: at (3,4) it gives minus three quarters, the negative reciprocal of the radius's slope of four thirds, so the tangent is perpendicular to the radius as it must be.
OpenStax Calculus Volume 1, §3.8 Implicit Differentiation §3.8, pp. 269-270
Matching
Watch for a y.
Match the pairs
Why: The third needs the product rule as well, and its second term still carries the implicit factor. Any term containing a y produces one, whatever else is going on in that term.
Worked example
Checkpoint 3.60. The product rule and the chain rule together.
\[ \text{Find } \frac{dy}{dx} \text{ for } x^{2}y + y^{3} = 8. \]
Differentiate the first term with the product rule
Why: It is x squared times y.
\[ 2 x y + x ^{2}(\,dy / \,dx) \]
Differentiate the second term
Why: Chain rule on the cube.
\[ 3 y ^{2}(\,dy / \,dx) \]
Set the sum equal to zero
Why: The right side is a constant.
\[ 2 x y + x ^{2} y' + 3 y ^{2} y' = 0 \]
Collect the y prime terms
Why: Factor it out.
\[ y'(x ^{2} + 3 y ^{2}) = -2 x y \]
Solve
Why: Divide.
\[ y' = -2 x y / (x ^{2} + 3 y ^{2}) \]
Figure (svg): The solution to Worked example a product of x and y shown as a ladder of expressions, one row per legal move
\[ \frac{dy}{dx} = \frac{-2xy}{x^{2}+3y^{2}} \]
Verify: check the term that needed two rules
Why: The first term x squared y is a product in which the second factor also carries a y, so both the product rule AND the chain rule apply: 2x times y plus x squared times y prime. Getting only one of the two is the standard failure. As a check, at the point (0, 2) — which satisfies the equation since 8 equals 8 — the derivative is 0, and the curve does have a horizontal tangent where it crosses the vertical axis.
OpenStax Calculus Volume 1, §3.8 Implicit Differentiation §3.8, pp. 270-271
Trap
\[ x^{2}+y^{2}=25 \]
Differentiate y squared as though y were the variable
Why: The student writes 2y and stops.
\[ 2x + 2y = 0 \;\Longrightarrow\; y = -x \quad \text{(nonsense)} \]
The result is an equation with no derivative in it at all, and it describes a line rather than a circle's slope.
\[ 2x + 2y\frac{dy}{dx} = 0 \;\Longrightarrow\; \frac{dy}{dx} = -\frac{x}{y} \]
Attach dy by dx to every differentiated y
Why: y is a function of x, so the chain rule applies to every y term.
The symptom is unmistakable: if no dy by dx survives the differentiation, the factor was dropped and there is nothing left to solve for. Writing y as f of x while differentiating makes the composition visible and the factor unavoidable.
Fill the middle
Differentiating the circle's equation term by term.
Fill in the blanks
2x + 2y\,\frac{dy}{dx} = 0
Why: Differentiating y squared with respect to x gives 2y times dy by dx, because y is a function of x. Without that factor there is nothing to solve for and the method collapses.
Sorting
Check whether the term contains a y.
Sort into buckets
Sort each term, differentiated with respect to x.
The fourth is the one to watch: it contains a y AND needs the product rule, so it produces one term without the factor and one with it. Scanning each term for a y before differentiating it is the discipline that makes this reliable.
Prediction
Commit before reasoning.
Predict first
Why does differentiating y squared with respect to x give 2y times dy/dx rather than 2y?
Correct: Because y is a function of x, so y squared is a composition.
\[ \frac{d}{dx}\left[\big(f(x)\big)^{2}\right] = 2f(x)f'(x) \]
Why: Writing y as f of x makes it plain: y squared is f of x, all squared, which is the outer squaring applied to the inner function f. The chain rule then gives 2 f of x times f prime of x, and f prime of x is what dy by dx names. Nothing is conventional or convenient about it — the factor is required by a rule proved in Section 3.6, and omitting it produces an equation with no derivative in it at all.
Section
Section 2
Concept
After differentiating, the equation contains dy by dx in one or more terms. Gather those on one side, factor the derivative out, and divide. The answer will generally involve both x and y.
an implicit derivative — A formula for dy by dx in terms of both x and y. It gives the slope at any point ON the curve, and both coordinates of that point are needed to evaluate it.
\[ y'\big(\text{stuff}\big) = \text{other stuff} \;\Longrightarrow\; y' = \frac{\text{other stuff}}{\text{stuff}} \]
An answer naming both variables is not incomplete. It is the correct form, because a point on an implicit curve is specified by two coordinates and the slope genuinely depends on both.
Figure (svg): The four steps of implicit differentiation, applied to a circle
OpenStax Calculus Volume 1, §3.8 Implicit Differentiation §3.8, pp. 269-272 — solving for the derivative
Picture it
Differentiate, collect, solve.
Figure (svg): The four steps of implicit differentiation, applied to a circle
The final line names both variables, and the note beneath is the point: that is the answer's proper form, not a sign that more work remains.
Worked example
Example 3.62. Collect carefully.
\[ \text{Find } \frac{dy}{dx} \text{ for } x^{3}+y^{3}=6xy. \]
Differentiate the left side
Why: Chain rule on the y cube.
\[ 3 x ^{2} + 3 y ^{2} y' \]
Differentiate the right side with the product rule
Why: Six times a product of x and y.
\[ 6 y + 6 x y' \]
Set them equal and move the y prime terms together
Why: Collect.
\[ 3 y ^{2} y' - 6 x y' = 6 y - 3 x ^{2} \]
Factor out the derivative
Why: Common factor.
\[ y'(3 y ^{2} - 6 x) = 6 y - 3 x ^{2} \]
Solve and simplify
Why: Divide by 3.
\[ y' = \frac{2 y - x ^{2}}{y ^{2} - 2 x} \]
Figure (svg): The folium of Descartes, a curve with no function description at all
\[ \frac{dy}{dx} = \frac{2y - x^{2}}{y^{2} - 2x} \]
Verify: evaluate at a point known to be on the curve
Why: The point (3, 3) satisfies the equation: 27 plus 27 is 54, and 6 times 9 is also 54. Substituting gives (6 minus 9) over (9 minus 6), which is negative 1 — so the tangent there has slope negative 1. The folium is symmetric about the line y equals x, and (3, 3) sits on that line, so a tangent perpendicular to it is exactly right. Note that solving this cubic for y would require the cubic formula and produce three branches; the implicit method never needed to.
OpenStax Calculus Volume 1, §3.8 Implicit Differentiation §3.8, pp. 271-272
Ranking
Implicit differentiation from start to finish.
Put in order
Why: Step d is the one that goes wrong: the derivative must come out of every term at once, as a single factor, before any division. Dividing term by term produces stray pieces outside the fraction.
Worked example
Checkpoint 3.62. Both sides carry the factor.
\[ \text{Find } \frac{dy}{dx} \text{ for } \sin(xy) = x. \]
Differentiate the left side
Why: Chain rule outside, product rule inside.
\[ \cos(x y) (y + x y') \]
Differentiate the right side
Why: The identity.
\[ 1 \]
Expand the left side
Why: Distribute the cosine.
\[ y \cos(x y) + x y' \cos(x y) = 1 \]
Isolate the y prime term
Why: Subtract.
\[ x y' \cos(x y) = 1 - y \cos(x y) \]
Solve
Why: Divide.
\[ y' = \frac{1 - y \cos(x y)}{x \cos(x y)} \]
Figure (svg): The solution to Worked example a derivative appearing twice shown as a ladder of expressions, one row per legal move
\[ \frac{dy}{dx} = \frac{1 - y\cos(xy)}{x\cos(xy)} \]
Verify: confirm no explicit solution was ever possible
Why: This equation cannot be solved for y in any closed form — there is no formula for y in terms of x. Yet the derivative was obtained in five lines. That is the strongest case for the technique: it is not merely more convenient than solving, it works where solving is impossible. Note also that the derivative fails where cosine of xy vanishes, which is exactly where the curve has vertical tangents.
OpenStax Calculus Volume 1, §3.8 Implicit Differentiation §3.8, pp. 272-272
Error analysis
A student solves for the derivative on a curve.
Annotate
On: \( 3y^{2}y' - 6xy' = 6y - 3x^{2} \;\Longrightarrow\; y' = \frac{6y - 3x^{2}}{3y^{2}} - 6x \)
Factoring the derivative out of every term that contains it, as one step, before dividing, is what keeps this correct. The symptom is an answer with a stray term hanging outside the fraction.
Fill the middle
The folium's differentiated equation, with the y prime terms gathered.
Fill in the blanks
y'\big(3y^y^2 - 2x - 6x\big) = 6y - 3x^___ \;\Longrightarrow\; y' = \frac___}___}
Why: Dividing both parts by 3 gives the tidier form. The derivative involves both x and y, which is the normal shape of an implicit answer.
Two truths and a lie
All three are about the answers this method gives.
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. For most implicit curves there is no way to eliminate y, and there is no need: to evaluate the slope you need a point on the curve, and such a point supplies both coordinates. Demanding an answer in x alone would make the method useless for exactly the curves it exists to handle.
Prediction
Commit before reasoning.
Predict first
For which equation is implicit differentiation the only available approach?
Correct: The one with no closed-form solution for y.
\[ \sin(xy) = x: \quad \text{no } y = f(x) \text{ exists in closed form} \]
Why: The first and last are already solved or trivially solvable. The circle can be solved for y at the cost of splitting into two branches, so implicit differentiation is cleaner but not essential. The trigonometric equation genuinely cannot be solved for y in any finite formula, so there is no explicit route to fall back on — and the method still produces the derivative in five lines. That is where it stops being a convenience.
Section
Section 3
Concept
To find a tangent, substitute both coordinates of the point into the implicit derivative to get the slope, then use point-slope form. The curve need not be a function for this to work.
tangent to an implicit curve — The line through a point on the curve whose slope is the implicit derivative evaluated at that point's two coordinates. Both coordinates are required, since the derivative depends on both.
\[ y - y_{0} = \left.\frac{dy}{dx}\right|_{(x_{0},y_{0})}(x - x_{0}) \]
This is where the technique visibly outperforms the explicit approach. A circle is not a function of x, so no single formula y equals f of x covers it, and yet every point has a perfectly good tangent.
Figure (svg): A circle with a tangent at a point, showing a curve that is not a function of x
OpenStax Calculus Volume 1, §3.8 Implicit Differentiation §3.8, pp. 271-273 — tangent lines to implicit curves
Picture it
The circle at the point three comma four.
Figure (svg): A circle with a tangent at a point, showing a curve that is not a function of x
The dashed radius has slope four thirds and the tangent minus three quarters — perpendicular, which is a fact about circles the derivative reproduces automatically.
Worked example
Example 3.63. Both coordinates go into the slope.
\[ \text{Find the tangent to } x^{2}+y^{2}=25 \text{ at } (3,4). \]
Check the point is on the curve
Why: Nine plus 16.
\[ 25,\text{ so yes} \]
Use the implicit derivative
Why: From the earlier example.
\[ \,dy / \,dx = -\frac{x}{y} \]
Substitute both coordinates
Why: Three over 4, negated.
\[ \text{slope } = -\frac{3}{4} \]
Write point-slope form
Why: Point and slope.
\[ y - 4 = -(\frac{3}{4}) (x - 3) \]
Rearrange
Why: Distribute and collect.
\[ y = -(\frac{3}{4}) x + \frac{25}{4} \]
Figure (svg): The solution to Worked example a tangent to a circle shown as a ladder of expressions, one row per legal move
\[ y - 4 = -\tfrac{3}{4}(x-3) \]
Verify: check perpendicularity to the radius
Why: The radius from the origin to (3,4) has slope four thirds, and the tangent's slope is minus three quarters — their product is negative 1, so they are perpendicular. That is a defining property of a circle's tangent, and it emerged from the calculus without being assumed. Checking the point lies on the curve first, as step one did, is essential: the derivative formula gives a slope for any pair of numbers, including pairs that are nowhere near the circle.
OpenStax Calculus Volume 1, §3.8 Implicit Differentiation §3.8, pp. 272-273
Fill the middle
The circle's derivative, evaluated at the point three comma four.
Fill in the blanks
\left.-\frac-3/4___\right|____ = ___
Why: Substituting x equal to 3 and y equal to 4 gives minus three quarters. At the point (3, -4) the same formula gives plus three quarters, which is why both coordinates are required.
Worked example
Checkpoint 3.63. Same method on a much harder curve.
\[ \text{Find the tangent to } x^{3}+y^{3}=6xy \text{ at } (3,3). \]
Confirm the point is on the curve
Why: Twenty-seven plus 27 against 6 times 9.
\[ 54 = 54 \]
Use the implicit derivative
Why: From the earlier example.
\[ \frac{2 y - x ^{2}}{y ^{2} - 2 x} \]
Substitute both coordinates
Why: Six minus 9, over 9 minus 6.
\[ -\frac{3}{3} = -1 \]
Write the tangent
Why: Point-slope with slope negative 1.
\[ y - 3 = -(x - 3) \]
Simplify
Why: Collect.
\[ y = -x + 6 \]
Figure (svg): The solution to Worked example a tangent to the folium shown as a ladder of expressions, one row per legal move
\[ y = 6 - x \]
Verify: use the curve's symmetry
Why: The folium is symmetric about the line y equals x, since swapping the variables leaves the equation unchanged. The point (3,3) lies on that line of symmetry, so the tangent there must be perpendicular to it — slope negative 1, which is what the derivative gave. A symmetry check like this is often the only independent verification available for an implicit curve, since there is no explicit formula to differentiate a second way.
OpenStax Calculus Volume 1, §3.8 Implicit Differentiation §3.8, pp. 273-273
Trap
\[ \frac{dy}{dx} = -\frac{x}{y} \text{ at } (3,4) \]
Substitute x equal to 3 and stop
Why: The student treats the derivative as a function of x alone.
\[ \text{slope} = -\frac{3}{y} \quad \text{(not a number)} \]
The answer still contains y, so it is not a slope at all and cannot be used in point-slope form.
\[ \left.-\frac{x}{y}\right|_{(3,4)} = -\frac{3}{4} \]
Substitute BOTH coordinates of the point
Why: An implicit derivative is a function of two variables, and a point supplies both.
This is the practical consequence of the answer's form. A point on an implicit curve is a pair, and the slope genuinely differs between points sharing an x-coordinate — on the circle, (3,4) and (3,-4) have slopes minus three quarters and plus three quarters.
Matching
The derivative is minus x over y.
Match the pairs
Why: The first two share an x-coordinate and have opposite slopes, which is exactly why both coordinates are needed. The last two are the horizontal and vertical tangents, and they are the subject of the next idea.
Ranking
Finding a tangent to an implicit curve.
Put in order
Why: Step a is worth doing first because the derivative formula will happily produce a slope for a point nowhere near the curve, and the resulting tangent line would touch nothing. Step d is where both coordinates are needed.
Prediction
Commit before reasoning.
Predict first
A circle fails the vertical line test, so it is not a function. How can it have a tangent at every point?
Correct: Because tangency is local, and near any point the circle is locally a function.
\[ \text{near } (5,0): \; x = \sqrt{25-y^{2}} \text{ is a function of } y \]
Why: Being a function is a global condition about the whole curve; having a tangent is a statement about the immediate neighbourhood of one point. Near (3,4) the circle is the graph of the upper semicircle, a perfectly good function; near (5,0) it is not a function of x but is a function of y. The implicit method quietly handles both cases without ever choosing a branch, which is precisely why it is more convenient than solving even when solving is possible.
Section
Section 4
Concept
An implicit derivative is a fraction. Horizontal tangents occur where its numerator vanishes and its denominator does not; vertical tangents occur where the denominator vanishes and the numerator does not.
locating special tangents — For an implicit derivative written as a quotient, the points on the curve where the numerator is zero give horizontal tangents, and those where the denominator is zero give vertical ones. Both conditions must be combined with the original equation.
\[ \frac{dy}{dx} = \frac{N}{D}: \quad N = 0 \Rightarrow \text{horizontal}, \quad D = 0 \Rightarrow \text{vertical} \]
The condition must be solved together with the original equation, since only points actually on the curve count. A numerator vanishing at a point not on the curve says nothing at all.
Figure (svg): The circle's implicit derivative, showing where it is zero and where it is undefined
OpenStax Calculus Volume 1, §3.8 Implicit Differentiation §3.8, pp. 272-274 — horizontal and vertical tangents
Picture it
Where the derivative vanishes and where it fails.
Figure (svg): The circle's implicit derivative, showing where it is zero and where it is undefined
The top and bottom of the circle have horizontal tangents, where x is zero; the left and right extremes have vertical ones, where y is zero. One fraction locates both.
Worked example
Example 3.64. Solve each condition with the equation.
\[ \text{Find all horizontal and vertical tangents to } x^{2}+y^{2}=25. \]
Write the derivative
Why: From the earlier example.
\[ \,dy / \,dx = -\frac{x}{y} \]
Set the numerator to zero for horizontal tangents
Why: x equal to 0.
\[ x = 0 \]
Combine with the original equation
Why: Substitute to find y.
\[ y ^{2} = 25,\text{ so } y = +- 5 \]
Set the denominator to zero for vertical tangents
Why: y equal to 0.
\[ y = 0 \]
Combine again
Why: Substitute.
\[ x = +- 5 \]
Figure (svg): The solution to Worked example special tangents on a circle shown as a ladder of expressions, one row per legal move
\[ \text{horizontal: } (0,\pm 5); \quad \text{vertical: } (\pm 5, 0) \]
Verify: check against the picture
Why: The circle's top and bottom are at (0, 5) and (0, negative 5), where the curve is momentarily level — horizontal tangents. Its left and right extremes are at (negative 5, 0) and (5, 0), where the curve is momentarily vertical. All four match. Note that each condition had to be solved TOGETHER with the original equation: x equal to 0 alone describes a whole line, and only its two intersections with the circle are points on the curve.
OpenStax Calculus Volume 1, §3.8 Implicit Differentiation §3.8, pp. 273-274
Sorting
For dy/dx = -x/y on the circle of radius 5.
Sort into buckets
Sort each point on the circle.
One fraction locates both kinds of special tangent, and reading its two parts separately is far quicker than examining the curve. The case where BOTH vanish is different again — it signals a self-intersection or a cusp, as at the folium's origin.
Worked example
Checkpoint 3.64. The same conditions on a harder curve.
\[ \text{Find the horizontal tangents to } x^{3}+y^{3}=6xy. \]
Write the derivative
Why: From the earlier example.
\[ \frac{2 y - x ^{2}}{y ^{2} - 2 x} \]
Set the numerator to zero
Why: Horizontal tangent condition.
\[ 2 y = x ^{2},\text{ so } y = x ^{2} / 2 \]
Substitute into the original equation
Why: Replace y.
\[ x ^{3} + x ^{6} / 8 = 3 x ^{3} \]
Solve
Why: Collect and factor.
\[ x ^{6} = 16 x ^{3},\text{ so } x = 0\text{ or } x ^{3} = 16 \]
Find the corresponding y and discard the degenerate point
Why: At x equal to 0 the denominator also vanishes.
\[ x = 16 ^{\frac{1}{3}}, y = x ^{2} / 2 \]
Figure (svg): The folium of Descartes, a curve with no function description at all
\[ x = 16^{1/3} \approx 2.52, \quad y = \tfrac{x^{2}}{2} \approx 3.17 \]
Verify: check the discarded point and the retained one
Why: At the origin BOTH the numerator and the denominator vanish, so the derivative is indeterminate there — the folium crosses itself at the origin and has two different tangents, so no single slope exists. Discarding it was necessary. The retained point, at about (2.52, 3.17), sits at the top of the loop, which is exactly where a horizontal tangent belongs. Checking each candidate against the denominator is what separates genuine horizontal tangents from indeterminate points.
OpenStax Calculus Volume 1, §3.8 Implicit Differentiation §3.8, pp. 274-274
Error analysis
A student hunts for horizontal tangents on the circle.
Annotate
On: \( \frac{dy}{dx} = -\frac{x}{y} = 0 \;\Longrightarrow\; x = 0, \text{ so the tangent is horizontal along the whole } y\text{-axis} \)
A derivative formula is only meaningful at points that satisfy the original equation. Every condition extracted from it must be intersected with the curve before it names any points.
Fill the middle
Hunting horizontal tangents on the circle of radius 5.
Fill in the blanks
x = 0 \text\pm 5 x^___+y^___=25 \;\Longrightarrow\; y = ___
Why: Substituting x equal to 0 into the circle's equation gives y squared equal to 25, so y is plus or minus 5. The condition alone described a whole line; intersecting it with the curve gives the two actual points.
Two truths and a lie
All three are about special tangents.
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. A vanishing denominator gives a VERTICAL tangent, which is a perfectly good tangent line — it simply has no slope as a number. The circle at (5, 0) has the vertical line x equals 5 as its tangent. What genuinely signals trouble is both parts vanishing at once, as at the folium's self-intersection.
Prediction
Commit before reasoning.
Predict first
At the folium's origin both the numerator and the denominator of dy/dx vanish. What does that indicate?
Correct: A self-intersection, with two tangents and no single slope.
\[ \text{at } (0,0): \; \frac{2(0) - 0^{2}}{0^{2} - 2(0)} = \frac{0}{0} \]
Why: The origin satisfies the folium's equation, so it is certainly on the curve, and the derivative is indeterminate there — zero over zero, which as Section 2.3 established determines nothing. Geometrically the curve passes through the origin twice, along two different directions, so no single tangent exists. This is why every candidate point must be checked against BOTH parts of the fraction rather than just the one being set to zero.
Section
Section 5
Concept
Differentiating the implicit derivative produces a new expression containing dy by dx. Substituting the known first derivative removes it, and the original equation is then often used to simplify further.
implicit second derivative — Obtained by differentiating the first derivative with respect to x, substituting the first derivative wherever it reappears, and simplifying with the original equation.
\[ \frac{d^{2}y}{dx^{2}} \text{ from } \frac{dy}{dx}, \text{ by differentiating and substituting} \]
Two substitutions are involved, and they happen at different moments. The first derivative goes in as soon as it appears; the original equation is used at the end, to tidy.
Figure (svg): Finding a second derivative implicitly, with the substitution step highlighted
OpenStax Calculus Volume 1, §3.8 Implicit Differentiation §3.8, pp. 273-274 — higher derivatives implicitly
Picture it
The circle's second derivative.
Figure (svg): Finding a second derivative implicitly, with the substitution step highlighted
The middle line is where the first derivative reappears and must be substituted away; the last is where the original equation collapses the expression to something remarkably simple.
Worked example
Example 3.65. Quotient rule, then substitute.
\[ \text{Find } \frac{d^{2}y}{dx^{2}} \text{ for } x^{2}+y^{2}=25. \]
Start from the first derivative
Why: Known already.
\[ \,dy / \,dx = -\frac{x}{y} \]
Differentiate with the quotient rule
Why: Remembering the implicit factor on y.
\[ -(y - x(\,dy / \,dx)) / y ^{2} \]
Substitute the first derivative
Why: Replace dy/dx by minus x over y.
\[ -(y + x ^{2} / y) / y ^{2} \]
Combine over a common denominator
Why: Multiply through by y.
\[ -(y ^{2} + x ^{2}) / y ^{3} \]
Use the original equation
Why: The numerator is 25.
\[ -25 / y ^{3} \]
Figure (svg): The solution to Worked example the circle's second derivative shown as a ladder of expressions, one row per legal move
\[ \frac{d^{2}y}{dx^{2}} = -\frac{25}{y^{3}} \]
Verify: check the sign on each half of the circle
Why: On the upper half y is positive, so the second derivative is negative and the curve bends downward — which the upper semicircle does. On the lower half y is negative, so y cubed is negative and the second derivative is positive, meaning the curve bends upward — which the lower semicircle does. Both match the picture. Note how the original equation collapsed x squared plus y squared into 25 at the last step; without that the answer would be correct but far uglier.
OpenStax Calculus Volume 1, §3.8 Implicit Differentiation §3.8, pp. 274-274
Fill the middle
The circle's second derivative, after the first derivative has been substituted.
Fill in the blanks
-\frac25+x^___}___} = -\frac___}___}
Why: The original equation says x squared plus y squared is 25, so the numerator collapses to a constant. Using the equation at the end is what turns a messy expression into a simple one.
Worked example
Checkpoint 3.65. The second derivative describes the bending.
\[ \text{Use } \frac{d^{2}y}{dx^{2}} = -\frac{25}{y^{3}} \text{ to describe the circle's curvature.} \]
Examine the sign on the upper half
Why: y positive, so y cubed positive.
Interpret
Why: A negative second derivative means bending downward.
Examine the lower half
Why: y negative, so y cubed negative.
Interpret
Why: Bending upward.
Note the magnitude near the extremes
Why: As y approaches 0 the expression is unbounded.
\[ \text{steepest bending near } (+- 5, 0) \]
Figure (svg): The solution to Worked example interpreting the result shown as a ladder of expressions, one row per legal move
\[ y > 0 \Rightarrow \text{concave down}; \quad y < 0 \Rightarrow \text{concave up} \]
Verify: ask whether the unbounded value is a problem
Why: As y approaches 0 the second derivative grows without bound, which sounds alarming but is exactly right: those are the points with vertical tangents, where y as a function of x is not differentiable at all. The circle itself is perfectly smooth there — it is the DESCRIPTION of it as a function of x that fails. This is a good reminder that an implicit derivative describes y as a function of x, and inherits that description's limitations.
OpenStax Calculus Volume 1, §3.8 Implicit Differentiation §3.8, pp. 274-274
Trap
\[ \frac{d^{2}y}{dx^{2}} = -\frac{y - x\frac{dy}{dx}}{y^{2}} \]
Stop after the quotient rule
Why: The student treats this as a finished answer.
The expression still contains the first derivative, so it cannot be evaluated at a point without computing that separately.
\[ \text{substitute } \frac{dy}{dx} = -\frac{x}{y} \;\Longrightarrow\; -\frac{25}{y^{3}} \]
Substitute the known first derivative, then simplify with the original equation
Why: Both substitutions are part of the method, not optional tidying.
The payoff is dramatic here: an expression in x, y and y prime collapses to a single term in y alone. That collapse is typical, and it only happens if both substitutions are made — the first derivative, and then the original equation.
Ranking
Finding d squared y by dx squared implicitly.
Put in order
Why: Steps c and e are two different substitutions at two different moments, and both are needed. Omitting c leaves an unevaluable answer; omitting e leaves a correct but needlessly complicated one.
Sorting
Using the second derivative, minus 25 over y cubed.
Sort into buckets
Sort each point on the circle of radius 5.
The sign depends only on which half the point is on, not on x at all — which matches the picture, since the upper arc curves one way throughout and the lower arc the other. Chapter 4 will call these concave down and concave up and use them to classify turning points.
Prediction
Commit before reasoning.
Predict first
Why does the circle's second derivative collapse to a single term?
Correct: Because the original equation replaces x squared plus y squared with 25.
\[ -\frac{x^{2}+y^{2}}{y^{3}} = -\frac{25}{y^{3}} \quad \text{on this curve only} \]
Why: After combining over a common denominator the numerator is exactly x squared plus y squared, which the curve's own equation says is 25. Using the original equation as a final simplification is a standard and often dramatic step in implicit differentiation, and it is available precisely because every point being considered satisfies that equation. Symmetry is a real property of the circle but is not what produced this collapse.
Comparison
Fill the blanks. The right column is why the technique exists.
Comparison matrix
| Situation | Explicit route | Implicit route |
|---|---|---|
| y already solved for | differentiate directly | also works, but adds nothing |
| A circle | split into two branches first | one computation covers both halves |
| The folium | needs the cubic formula, three branches | five lines |
| sin(xy) = x | impossible | five lines, exactly as before |
Reading down the middle column shows the explicit route degrading from easy to impossible, while the implicit one stays the same length throughout. That constancy is the technique's real value.
Pattern
Given an equation in x and y and asked for the derivative.
For a second derivative, differentiate the first, substitute the first derivative wherever it reappears, and then use the original equation to simplify. Both substitutions are part of the method.
Stewart, Calculus: Early Transcendentals 8e, §3.5 Implicit Differentiation §3.5, pp. 208-217
Check
The implicit factor. Every y produces one.
Check your understanding
Differentiating x^2 + y^2 = 25 with respect to x gives what?
Answer: A
Why: The y term needs a chain-rule factor; the constant 25 differentiates to 0.
Check
Evaluating. Both coordinates.
Check your understanding
For dy/dx = -x/y on the circle, what is the slope at (3, -4)?
Answer: A
Why: Minus 3 over negative 4 is positive three quarters.
Check
Special tangents. Read the two parts.
Check your understanding
For dy/dx = -x/y on the circle, where are the vertical tangents?
Answer: A
Why: The denominator vanishes when y = 0, and combining with the equation gives x = plus or minus 5.
Real world
A gas in a sealed cylinder obeys the relation P V to the power 1.4 equals a constant, where P is pressure and V is volume. An engineer needs to know how sensitive the pressure is to a change in volume during compression.
Discussion prompt
Differentiate the relation implicitly with respect to V, find dP by dV, and say what the negative sign and the magnitude mean physically.
Hint: P is a function of V, so every P differentiated produces a dP by dV.
Answer:
\[ PV^{1.4} = C \;\Longrightarrow\; \frac{dP}{dV}V^{1.4} + P(1.4)V^{0.4} = 0 \]
Both terms came from the product rule, and the first carries the implicit factor because P depends on V. Solving:
\[ \frac{dP}{dV} = -\frac{1.4P}{V} \]
The negative sign says pressure falls as volume grows, which is what compressing and expanding a gas does — and it came out of the algebra rather than being assumed.
The magnitude is more interesting. The sensitivity is proportional to the current pressure and inversely proportional to the current volume, so a highly compressed gas responds far more sharply to a small volume change than a slack one does. At a tenth the volume and ten times the pressure, the same change in volume produces a hundred times the pressure change.
Note what was never needed: the constant C, and any attempt to solve for P explicitly. Solving would have been possible here, but the implicit route gave the answer in terms of the current state P and V — which is exactly the form an engineer wants, since those are the quantities a gauge reports.
Commit first
Answer, then rate your confidence honestly.
Predict first
What is the derivative of y squared with respect to x, when y is a function of x?
Correct: Two y times dy by dx.
\[ \frac{d}{dx}\left[y^{2}\right] = 2y\frac{dy}{dx} \quad \text{always, when } y = y(x) \]
Why: Because y depends on x, y squared is a composition and the chain rule attaches the factor dy by dx. Writing y as f of x makes it obvious: the derivative of f of x squared is 2 f of x times f prime of x. Omitting the factor is the error that destroys the whole method, because it leaves an equation with no derivative in it to solve for. The answer 2x would be right only if the term were x squared, which is a different term entirely.
Explain it
They differentiated the circle's equation and got 2x plus 2y equals 0, and cannot see what went wrong.
Discussion prompt
In four sentences or fewer, show them the error without quoting the rule.
Hint: Have them rewrite y as f of x first.
Answer:
Ask them to write y as f of x everywhere and try again: the equation becomes x squared plus f of x squared equals 25. Now the second term is visibly a composition, so the chain rule applies and gives 2 f of x times f prime of x.
Translating back, that is 2y times dy by dx — the factor they dropped. And notice what their version produced: an equation with no derivative in it at all, which is a sure sign the factor went missing, since the whole point was to solve for one.
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 factor, mentally rewrite y as f of x before differentiating any term. For collecting, factor the derivative out of every term in one step before dividing. For evaluating, remember an implicit derivative needs a point, not an input. For second derivatives, substitute the first derivative as soon as it reappears and use the original equation at the end. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, write an x term and a y term side by side, differentiate each with respect to x, and circle the extra factor the y term produces, with one sentence saying where it comes from. Below, take the circle of radius 5 and run the whole method: differentiate, collect, solve, then draw the circle and mark the tangent at the point three comma four, checking it is perpendicular to the radius. On the same circle mark the two horizontal and two vertical tangents, and beside each write which part of the fraction vanishes there. In the middle of the page, differentiate the folium implicitly in full, and evaluate the slope at three comma three, noting the symmetry that confirms it. At the bottom, compute the circle's second derivative completely, marking the two substitutions — the first derivative, and then the original equation — and write one sentence on what its sign says about each half of the circle. In a margin, write an equation that cannot be solved for y at all, and one sentence on why the method still works.
If your differentiated circle equation has no dy by dx in it anywhere, the implicit factor was dropped — there would then be nothing to solve for, which is the symptom to recognise.
Recap
Five things, and the technique works on curves no function can describe.
| If you see | Then |
|---|---|
| A term containing y | Its derivative carries a dy/dx |
| A term like x^2 y | Product rule AND chain rule |
| No dy/dx after differentiating | The factor was dropped |
| An answer with both x and y | That is the correct form |
| A zero numerator | A horizontal tangent, if the denominator is not also zero |
| A zero denominator | A vertical tangent |
| Both zero | A self-intersection or cusp: no single slope |
Section 3.9 completes Chapter 3 with the exponential and logarithmic families, and it uses this section's technique twice: once to differentiate the logarithm, and once for logarithmic differentiation, which turns awkward products and powers into sums.
OpenStax Calculus Volume 1, §3.8 Implicit Differentiation §3.8, pp. 268-274 — everything on these slides traces back here
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