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Lesson 54 — Implicit Differentiation

Differentiate y defined implicitly by equation F(x, y) = 0. Chain rule, tangent to implicit curves, implicit second derivative.

Used in: Japanese Equiv. Math III (implicit + inverse functions) · German Klasse 11 LK Equiv. · Singapore H2 Math (derivatives of curves)

\frac{d}{dx}\bigl[F(x,y)\bigr] = 0 \;\Longrightarrow\; \frac{dy}{dx} = -\frac{\partial F/\partial x}{\partial F/\partial y}

The implicit derivative of $F(x, y) = 0$ : differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ (chain rule), and isolate $dy/dx$ . The formula $-F_x/F_y$ holds where $F_y \neq 0$ .

Choose your door

Rigorous notation, full derivation, hypotheses

Definition and Implicit Function Theorem

Motivation

A plane curve can be given by $F(x, y) = 0$ without it being possible, or convenient, to isolate $y$ explicitly. The circle $x^2 + y^2 = r^2$ and the Folium of Descartes $x^3 + y^3 = 3axy$ are canonical examples. Implicit differentiation bypasses the obstacle.

Formal Recipe

Let $F(x, y) = 0$ be an equation defining $y$ as a function of $x$ in a neighborhood of a point $(a, b)$ .

Canonical Example: Circle

$x^2 + y^2 = r^2$

Differentiating: $2x + 2y\,y' = 0$ , whence $y' = -\dfrac{x}{y}$ (valid for $y \neq 0$ ).

Table of Classic Curves

Curve	Equation $F(x,y)=0$	$dy/dx$
Circle	$x^2 + y^2 - r^2 = 0$	$-x/y$
Ellipse	$x^2/a^2 + y^2/b^2 - 1 = 0$	$-(b^2 x)/(a^2 y)$
Hyperbola	$x^2/a^2 - y^2/b^2 - 1 = 0$	$(b^2 x)/(a^2 y)$
Folium of Descartes	$x^3 + y^3 - 3axy = 0$	$(ay - x^2)/(y^2 - ax)$

"If the equation relating $x$ and $y$ cannot be solved for $y$ explicitly, we can still find $y'$ by differentiating the equation implicitly." — OpenStax Calculus Volume 1, §3.8

Implicit Function Theorem (1D version)

When it fails. If $F_y(a, b) = 0$ , the curve may have a vertical tangent at that point, or it may not define a function locally. Example: the circle at points $(\pm r, 0)$ — $F_y = 2y = 0$ there.

Implicit Second Derivative

Apply $\tfrac{d}{dx}$ again to $y' = -F_x/F_y$ , using the quotient rule and remembering that $y$ depends on $x$ .

Solved Examples

Example— 1· Implicit Differentiation of a Circle (Foundation)

Problem. Find $dy/dx$ for the circle $x^2 + y^2 = 25$ .

Strategy. This is the standard example for implicit differentiation. Differentiate directly with respect to $x$ , remembering that $y$ is a function of $x$ .

Solution.

Differentiate both sides with respect to $x$ :

$\frac{d}{dx}\bigl[x^2 + y^2\bigr] = \frac{d}{dx}[25]$

Apply the chain rule to $y^2$ :

$2x + 2y\,\frac{dy}{dx} = 0$

Isolate $dy/dx$ :

$2y\,\frac{dy}{dx} = -2x \implies \frac{dy}{dx} = -\frac{x}{y}$

Check at point $(3, 4)$ : The point is on the curve ( $9 + 16 = 25$ ). The slope is $dy/dx = -3/4$ . The tangent line is $y - 4 = -\tfrac{3}{4}(x - 3)$ , or $3x + 4y = 25$ . Geometrically: the radius to $(3,4)$ has slope $4/3$ , and $(-3/4)(4/3) = -1$ , confirming tangent and radius are perpendicular.

Source. OpenStax Calculus Volume 1 §3.8, example 3.68, p. 318 — CC-BY-NC-SA 4.0 license.

Example— 2· Implicit Differentiation of an Ellipse — Tangent at a Specific Point

Problem. For the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ , find $dy/dx$ and the equation of the tangent line at $\left(2, \tfrac{3\sqrt{3}}{2}\right)$ .

Strategy. Apply implicit differentiation, then evaluate at the given point.

Solution.

Differentiating:

$\frac{d}{dx}\left[\frac{x^2}{16} + \frac{y^2}{9}\right] = 0 \implies \frac{2x}{16} + \frac{2y}{9}\,y' = 0$

Simplifying:

$\frac{x}{8} + \frac{2y}{9}\,y' = 0 \implies y' = -\frac{9x}{16y}$

At the point $\left(2, \tfrac{3\sqrt{3}}{2}\right)$ :

$y' = -\frac{9 \cdot 2}{16 \cdot \tfrac{3\sqrt{3}}{2}} = -\frac{18}{24\sqrt{3}} = -\frac{3}{4\sqrt{3}} = -\frac{\sqrt{3}}{4}$

Equation of the tangent line:

$y - \frac{3\sqrt{3}}{2} = -\frac{\sqrt{3}}{4}\left(x - 2\right) \implies y = -\frac{\sqrt{3}}{4}\,x + 2\sqrt{3}$

Check: Substituting $x = 2$ : $y = -\sqrt{3}/2 + 2\sqrt{3} = 3\sqrt{3}/2$ . Correct.

Source. Active Calculus 2.0 §2.7, Preview Activity 2.7.1, p. 154 — CC-BY-NC-SA 4.0 license.

Example— 3· Folium of Descartes — Implicit Differentiation with Product Rule

Problem. Find $dy/dx$ for the Folium of Descartes $x^3 + y^3 = 3xy$ .

Strategy. The right side $3xy$ requires the product rule when differentiating. Then, group terms and isolate $y'$ .

Solution.

Differentiating both sides with respect to $x$ :

$3x^2 + 3y^2\,y' = 3\,(y + x\,y')$

(On the right side: $\dfrac{d}{dx}[3xy] = 3\left[(1)\cdot y + x \cdot y'\right] = 3y + 3x\,y'$ .)

Expand and group terms with $y'$ :

$3x^2 + 3y^2\,y' = 3y + 3x\,y'$

$3y^2\,y' - 3x\,y' = 3y - 3x^2$

$y'(3y^2 - 3x) = 3(y - x^2)$

Isolate $y'$ :

$y' = \frac{3(y - x^2)}{3(y^2 - x)} = \frac{y - x^2}{y^2 - x}$

Check: At the point $\left(\tfrac{3}{2}, \tfrac{3}{2}\right)$ (which lies on the Folium with $a=1$ ):

$y' = \frac{3/2 - 9/4}{9/4 - 3/2} = \frac{-3/4}{3/4} = -1$

The tangent has slope $-1$ , which makes sense due to the curve's symmetry about the line $y = x$ at that point.

Source. OpenStax Calculus Volume 1 §3.8, exercise 175, p. 325 — CC-BY-NC-SA 4.0 license.

Example— 4· Implicit Second Derivative

Problem. For $x^2 + y^2 = r^2$ , find $d^2y/dx^2$ in terms of $x$ , $y$ , and $r$ .

Strategy. We already know $y' = -x/y$ . Differentiate one more time with respect to $x$ , applying the quotient rule and substituting $y'$ into the resulting expression.

Solution.

Start with $y' = -\dfrac{x}{y}$ .
Apply $\dfrac{d}{dx}$ using the quotient rule:

$y'' = -\frac{(x)'\,y - x\,(y)'}{y^2} = -\frac{y - x\,y'}{y^2}$

Substitute $y' = -x/y$ :

$y'' = -\frac{y - x \cdot (-x/y)}{y^2} = -\frac{y + x^2/y}{y^2} = -\frac{y^2 + x^2}{y^3}$

Use $x^2 + y^2 = r^2$ :

$y'' = -\frac{r^2}{y^3}$

Interpretation. For $y > 0$ (upper part of the circle), $y'' = -r^2/y^3 < 0$ — the curve is concave down, which is geometrically correct for the upper semicircle.

Source. Active Calculus 2.0 §2.7, Activity 2.7.3, p. 158 — CC-BY-NC-SA 4.0 license.

Example— 5· Logarithmic Derivative and Trigonometric Implicit Curve

Problem (two parts). (a) Use logarithmic differentiation to find $y'$ if $y = x^{\sin x}$ (with $x > 0$ ). (b) Find $dy/dx$ for $\sin(xy) = x - y$ . Evaluate at $(0, 0)$ .

Strategy. (a) Taking the logarithm before differentiating eliminates the variable exponent. (b) Implicit differentiation with the chain rule on $\sin(xy)$ requires the product rule in the argument.

Solution — part (a).

Take $\ln$ of both sides: $\ln y = \sin x \cdot \ln x$ .
Differentiate with respect to $x$ :

$\frac{y'}{y} = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x}$

Multiply by $y = x^{\sin x}$ :

$y' = x^{\sin x}\!\left(\cos x\,\ln x + \frac{\sin x}{x}\right)$

Solution — part (b).

Differentiating $\sin(xy) = x - y$ :

$\cos(xy) \cdot \frac{d}{dx}[xy] = 1 - y'$

$\cos(xy) \cdot (y + x\,y') = 1 - y'$

Expand and group:

$y\cos(xy) + x\cos(xy)\,y' = 1 - y'$

$y' \bigl(x\cos(xy) + 1\bigr) = 1 - y\cos(xy)$

$y' = \frac{1 - y\cos(xy)}{1 + x\cos(xy)}$

At $(0, 0)$ : $\cos(0 \cdot 0) = \cos 0 = 1$ .

$y'(0,0) = \frac{1 - 0 \cdot 1}{1 + 0 \cdot 1} = \frac{1}{1} = 1$

Check: Does $(0,0)$ satisfy $\sin(0 \cdot 0) = 0 - 0$ ? $\sin(0) = 0$ . Correct.

Source. APEX Calculus §2.6, examples 2.6.4 and 2.6.5, p. 117 — CC-BY-NC 4.0 license.

Exercise list

40 exercises · 10 with worked solution (25%)

Application 22Understanding 6Modeling 8Challenge 3Proof 1

Ex. 54.1Application
For the circle $x^2 + y^2 = 1$ , find $dy/dx$ .
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Differentiating $x^2 + y^2 = 1$ with respect to $x$ : $2x + 2y\,y' = 0$ . Isolating: $y' = -x/y$ . Valid where $y \neq 0$ .
Show step-by-step (with the why)
1. Write and differentiate. Starting with $x^2 + y^2 = 1$ , apply $d/dx$ to both sides: $d[x^2]/dx + d[y^2]/dx = 0$ .
2. Use chain rule on $y^2$ . Since $y$ is a function of $x$ , $d[y^2]/dx = 2y\,y'$ . Thus: $2x + 2y\,y' = 0$ .
3. Isolate $y'$ . Subtract $2x$ : $2y\,y' = -2x$ . Divide by $2y$ : $y' = -x/y$ .
Hint: The result $-x/y$ makes geometric sense — the tangent to the circle at a point $(x_0, y_0)$ is perpendicular to the radius $(x_0, y_0)$ , which has slope $y_0/x_0$ . The product of the slopes is $(-x_0/y_0)(y_0/x_0) = -1$ .
Ex. 54.2Application
For the ellipse $x^2/4 + y^2/9 = 1$ , calculate $dy/dx$ .
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For the ellipse $x^2/4 + y^2/9 = 1$ : differentiating, $x/2 + 2y\,y'/9 = 0$ , so $y' = -9x/(4y)$ .
Show step-by-step (with the why)
1. Differentiate each term. $d[x^2/4]/dx = x/2$ and $d[y^2/9]/dx = 2y\,y'/9$ . Equation: $x/2 + 2y\,y'/9 = 0$ .
2. Isolate $y'$ . $2y\,y'/9 = -x/2$ , then $y' = -9x/(4y)$ .
Note: Varying the semi-axes $a, b$ , the general formula for $x^2/a^2 + y^2/b^2 = 1$ is $y' = -b^2 x/(a^2 y)$ .
Ex. 54.3Application
For $xy = 1$ , calculate $dy/dx$ via implicit differentiation. Verify it matches differentiating $y = 1/x$ explicitly.
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For $xy = 1$ : differentiating (product rule), $y + x\,y' = 0$ , so $y' = -y/x$ . Check: $y = 1/x$ gives $y' = -1/x^2 = -(1/x)/x = -y/x$ — consistent.
Ex. 54.4Application
For the hyperbola $x^2/9 - y^2/16 = 1$ , calculate $dy/dx$ .
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For the hyperbola $x^2/9 - y^2/16 = 1$ : differentiating, $2x/9 - 2y\,y'/16 = 0$ , so $y' = 16x/(9y)$ . Compare with the ellipse: the sign is positive because the sign of the $y^2$ term is negative in the hyperbola equation.
Ex. 54.5ApplicationAnswer key
For $x^3 + y^3 = 6xy$ , calculate $dy/dx$ .
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For $x^3 + y^3 = 6xy$ : differentiating, $3x^2 + 3y^2\,y' = 6y + 6x\,y'$ . Grouping: $y'(3y^2 - 6x) = 6y - 3x^2$ , so $y' = (2y - x^2)/(y^2 - 2x)$ .
Ex. 54.6Application
For $x^2 - 2xy + 3y^2 = 1$ , calculate $dy/dx$ .
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For $x^2 - 2xy + 3y^2 = 1$ : $2x - 2(y + x\,y') + 6y\,y' = 0$ . Simplifying: $2x - 2y + y'(-2x + 6y) = 0$ , so $y' = (y - x)/(3y - x)$ .
Ex. 54.7Application
For $x^2 y + xy^2 = 6$ , calculate $dy/dx$ .
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For $x^2 y + xy^2 = 6$ , differentiating (product rule on each term): $(2x\,y + x^2\,y') + (y^2 + 2xy\,y') = 0$ . Grouping: $y'(x^2 + 2xy) = -(2xy + y^2)$ , so $y' = -(2xy + y^2)/(x^2 + 2xy)$ .
Show step-by-step (with the why)
1. Identify the terms. The equation is $x^2 y + x y^2 = 6$ . Each term is a product of functions of $x$ and $y(x)$ , so the product rule applies twice.
2. Differentiate the first term. $d[x^2 y]/dx = 2x \cdot y + x^2 \cdot y'$ (product: derivative of the first times the second, plus the first times the derivative of the second).
3. Differentiate the second term. $d[x y^2]/dx = 1 \cdot y^2 + x \cdot 2y\,y' = y^2 + 2xy\,y'$ .
4. Assemble the equation and isolate. $2xy + x^2\,y' + y^2 + 2xy\,y' = 0$ , therefore $y'(x^2 + 2xy) = -(2xy + y^2)$ and $y' = -(2xy + y^2)/(x^2 + 2xy)$ .
Hint: Factor the numerator and denominator: $y' = -y(2x + y)/(x(x + 2y))$ . We can check that at the point $(1, 2)$ , which satisfies $1\cdot2 + 1\cdot4=6$ , we have $y' = -2(2+2)/(1(1+4)) = -8/5$ .
Ex. 54.8Application
For $\tan y = x$ , calculate $dy/dx$ . Interpret the result as the derivative of $\arctan x$ .
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For $\tan y = x$ : differentiating, $\sec^2 y \cdot y' = 1$ , so $y' = 1/\sec^2 y = \cos^2 y$ . This exercise implicitly derives the $\arctan$ function: since $y = \arctan x$ , and $\cos^2(\arctan x) = 1/(1+x^2)$ , we recover the known formula $(\arctan x)' = 1/(1+x^2)$ .
Ex. 54.9ApplicationAnswer key
For $e^y = xy$ , calculate $dy/dx$ .
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For $e^y = xy$ : differentiating, $e^y\,y' = y + x\,y'$ . Grouping: $y'(e^y - x) = y$ , so $y' = y/(e^y - x)$ . Substituting $e^y = xy$ : $y' = y/(xy - x) = y/(x(y-1)) = 1/(x(1 - 1/y))$ . The cleanest form is to keep $y' = y/(e^y - x)$ .
Ex. 54.10ApplicationAnswer key
For $\ln(xy) = x + y$ , calculate $dy/dx$ .
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For $\ln(xy) = x + y$ : using $\ln(xy) = \ln x + \ln y$ , differentiate: $1/x + y'/y = 1 + y'$ . Rearranging: $y'/y - y' = 1 - 1/x$ , so $y'(1/y - 1) = (x-1)/x$ , and $y' = y(x-1)/(x(1-y))$ .
Ex. 54.11Application
For $\sqrt{x} + \sqrt{y} = 4$ , calculate $dy/dx$ and evaluate at the point $(1, 9)$ .
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For $\sqrt{x} + \sqrt{y} = 4$ : differentiating, $1/(2\sqrt{x}) + y'/(2\sqrt{y}) = 0$ , so $y' = -\sqrt{y}/\sqrt{x} = -\sqrt{y/x}$ . At the point $(1, 9)$ (check: $1 + 3 = 4$ ): $y' = -3$ .
Ex. 54.12Application
For $\cos(x + y) = y$ , calculate $dy/dx$ .
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For $\cos(x+y) = y$ : differentiating, $-\sin(x+y)(1 + y') = y'$ . Expand: $-\sin(x+y) - \sin(x+y)\,y' = y'$ . Group: $-\sin(x+y) = y'(1 + \sin(x+y))$ , so $y' = -\sin(x+y)/(1+\sin(x+y))$ .
Ex. 54.13ApplicationAnswer key
For $\sin(xy) = x$ , calculate $dy/dx$ .
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For $\sin(xy) = x$ : differentiating, $\cos(xy)(y + x\,y') = 1$ . Expand: $y\cos(xy) + x\cos(xy)\,y' = 1$ . Isolate: $y' = (1 - y\cos(xy))/(x\cos(xy) + 1)$ .
Ex. 54.14Application
For $y^3 + 3y = x$ , calculate $dy/dx$ and discuss whether the derivative exists at all points.
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For $y^3 + 3y = x$ : differentiating, $3y^2\,y' + 3\,y' = 1$ , so $y'(3y^2 + 3) = 1$ and $y' = 1/(3(y^2+1))$ . Note that $y^2 + 1 \geq 1 > 0$ for all $y$ , so $y'$ exists at every point on the curve — there are no vertical tangents.
Ex. 54.15ApplicationAnswer key
Find the tangent line to the circle $x^2 + y^2 = 25$ at the point $(3, 4)$ .
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For the circle $x^2+y^2=25$ , $y' = -x/y$ . At point $(3,4)$ : $y' = -3/4$ . Tangent line: $y - 4 = -3/4 (x-3)$ , or $4y - 16 = -3x + 9$ , so $3x + 4y = 25$ .
Show step-by-step (with the why)
1. Calculate $y'$ implicitly. Differentiating $x^2 + y^2 = 25$ : $2x + 2y\,y' = 0$ , therefore $y' = -x/y$ .
2. Evaluate at the point. At $(3,4)$ : check that $9 + 16 = 25$ . Yes. So $y' = -3/4$ .
3. Write the line. Point-slope form: $y - 4 = -3/4(x - 3)$ . Multiply by 4: $4y - 16 = -3x + 9$ , therefore $3x + 4y = 25$ .
4. Geometric check. The radius to $(3,4)$ has slope $4/3$ . Product $(-3/4)(4/3) = -1$ — tangent and radius are perpendicular.
Hint: For any circle centered at the origin, the tangent at $(x_0, y_0)$ has equation $x_0 x + y_0 y = r^2$ — an elegant result directly from implicit differentiation.
Ex. 54.16Application
For the ellipse $x^2 + 4y^2 = 16$ , find the tangent line at the point $(2, \sqrt{3})$ .
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For $x^2 + 4y^2 = 16$ : differentiating, $2x + 8y\,y' = 0$ , so $y' = -x/(4y)$ . At the point $(2, \sqrt{3})$ : check $4 + 4\cdot3 = 16$ . Correct. $y' = -2/(4\sqrt{3}) = -1/(2\sqrt{3}) = -\sqrt{3}/6$ . Tangent line: $y - \sqrt{3} = -\sqrt{3}/6\,(x-2)$ , or $y = -\sqrt{3}x/6 + \sqrt{3}/3 + \sqrt{3} = -\sqrt{3}x/6 + 4\sqrt{3}/3$ .
Ex. 54.17Application
For $x^2 + xy + y^2 = 7$ , find the tangent line at $(1, 2)$ .
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For $x^2 + xy + y^2 = 7$ : $2x + y + x\,y' + 2y\,y' = 0$ . At $(1,2)$ : check $1 + 2 + 4 = 7$ . Correct. $2 + 2 + y' + 4y' = 0 \Rightarrow 4 + 5y' = 0 \Rightarrow y' = -4/5$ . Tangent: $y - 2 = -4/5(x-1)$ , or $5y - 10 = -4x + 4$ , so $4x + 5y = 14$ .
Ex. 54.18Application
For $x^3 + y^3 = 9$ , find the tangent line at $(1, 2)$ .
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For $x^3 + y^3 = 9$ : $3x^2 + 3y^2\,y' = 0$ , $y' = -x^2/y^2$ . At $(1,2)$ : check $1 + 8 = 9$ . Correct. $y' = -1/4$ . Tangent: $y - 2 = -1/4(x-1) \Rightarrow y = -x/4 + 1/4 + 2 = -x/4 + 9/4$ .
Ex. 54.19Application
For $y\sin x = x\cos y$ , calculate $dy/dx$ .
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For $y\sin x = x\cos y$ : differentiating (product rule on both sides): $y'\sin x + y\cos x = \cos y - x\sin y\,y'$ . Group: $y'(\sin x + x\sin y) = \cos y - y\cos x$ , so $y' = (\cos y - y\cos x)/(\sin x + x\sin y)$ .
Ex. 54.20Application
For the circle $x^2 + y^2 = 1$ , determine all points of horizontal and vertical tangency.
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For the ellipse $x^2 + y^2 = 1$ (circle): $y' = -x/y$ . Vertical tangent when denominator $y = 0$ . Substituting $y=0$ into the curve: $x^2 = 1$ , so $x = \pm 1$ . The points of vertical tangency are $(1, 0)$ and $(-1, 0)$ . Horizontal tangent when numerator $x = 0$ : point $(0, \pm 1)$ .
Ex. 54.21Application
For the Folium of Descartes $x^3 + y^3 = 3xy$ , calculate $dy/dx$ and determine the points of horizontal tangency.
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For the Folium of Descartes $x^3 + y^3 = 3xy$ : $3x^2 + 3y^2\,y' = 3y + 3x\,y'$ . Grouping: $y'(3y^2 - 3x) = 3y - 3x^2$ , so $y' = (y - x^2)/(y^2 - x)$ . Horizontal tangent when numerator $y - x^2 = 0$ , i.e., $y = x^2$ . Substituting into the Folium: $x^3 + x^6 = 3x^3 \Rightarrow x^6 - 2x^3 = 0 \Rightarrow x^3(x^3-2)=0$ . Solutions: $x=0$ (singular) and $x = 2^{1/3}$ , with $y = 2^{2/3}$ .
Ex. 54.22Application
For the Folium of Descartes $x^3 + y^3 = 3xy$ , find the tangent at the point $(3/2, 3/2)$ .
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At the Folium $x^3 + y^3 = 3xy$ , the point $(3/2, 3/2)$ : check $2\cdot(27/8) = 27/4 = 3\cdot(3/2)\cdot(3/2) = 27/4$ . Yes. $y' = (3/2 - 9/4)/(9/4 - 3/2) = (-3/4)/(3/4) = -1$ . Tangent: $y - 3/2 = -(x - 3/2) \Rightarrow y = -x + 3$ .
Ex. 54.23Modeling
The ideal gas law states $PV = nRT$ . Holding $T$ constant, use implicit differentiation to find $dP/dV$ .
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From the ideal gas law $PV = nRT$ , with $T$ and $nR$ constant, differentiating implicitly with respect to $V$ : $\frac{dP}{dV}V + P = 0$ , so $dP/dV = -P/V = -nRT/V^2$ . This is the isothermal compressibility — the more compressed the gas, the faster pressure rises with further compression.
Show step-by-step (with the why)
1. Identify the variables. $P$ and $V$ are the variables; $nRT$ is constant (fixed temperature).
2. Differentiate both sides with respect to $V$ . $d[PV]/dV = d[nRT]/dV$ . Left side (product rule): $(dP/dV)V + P\cdot 1 = 0$ . Right side: zero.
3. Isolate $dP/dV$ . $(dP/dV)V = -P$ , so $dP/dV = -P/V$ . Substituting $P = nRT/V$ : $dP/dV = -nRT/V^2$ .
Curiosity: For the Van der Waals gas, $(P + a/V^2)(V - b) = nRT$ . Implicitly differentiating this more complex equation gives the real gas compressibility.
Ex. 54.24ModelingAnswer key
For the curve $y^2 + xy = 12$ , determine if there are any points of horizontal or vertical tangency.
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For $y^2 + xy - 12 = 0$ , $2y\,y' + y + x\,y' = 0$ , so $y' = -y/(2y+x)$ . Horizontal tangent when numerator $-y=0$ , i.e., $y=0$ . But $y=0$ on the curve implies $0 + 0 - 12 = 0$ — impossible. Vertical tangent when denominator $2y+x=0$ , i.e., $x=-2y$ . Substituting: $y^2 + (-2y)y - 12 = 0 \Rightarrow -y^2 = 12$ — impossible for real numbers. Therefore, there are no horizontal or vertical tangents at real points on this hyperbola.
Ex. 54.25Modeling
In microeconomics, the indifference curve $U(x_1, x_2) = \bar{U}$ describes combinations of two goods that leave the consumer indifferent. Using implicit differentiation, find $dx_2/dx_1$ — the marginal rate of substitution.
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An indifference curve is defined by $U(x_1, x_2) = \bar{U}$ (constant). Differentiating implicitly with respect to $x_1$ : $U_{x_1} + U_{x_2}\,(dx_2/dx_1) = 0$ , so $dx_2/dx_1 = -U_{x_1}/U_{x_2}$ . This is the marginal rate of substitution — the slope of the indifference curve. For a rational consumer, this rate is negative (if you want more of one good, you give up the other).
Ex. 54.26Modeling
For the lemniscate $(x^2+y^2)^2 = 2(x^2-y^2)$ , calculate $dy/dx$ at the point $(\sqrt{3}/2, 1/2)$ .
Solve online OpenStax Calculus Volume 1 · §3.8
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For the lemniscate $(x^2+y^2)^2 = 2(x^2-y^2)$ , differentiating both sides: $2(x^2+y^2)(2x+2y\,y') = 2(2x - 2y\,y')$ . Expand with $u = x^2 + y^2$ : $4u(x + y\,y') = 4x - 4y\,y'$ . Grouping $y'$ : $y'(4uy + 4y) = 4x - 4ux$ , so $y' = x(1-u)/(y(1+u))$ where $u = x^2+y^2$ . At point $(\sqrt{3}/2, 1/2)$ : $u = 3/4 + 1/4 = 1$ , therefore $y' = x(0)/(y(2)) = 0$ ... Reanalyze directly: with $u=1$ , $4(2x+2y\,y') = 4x - 4y\,y'$ , so $8x+8y\,y' = 4x-4y\,y'$ , $y'(8y+4y)=4x-8x=-4x$ , $y' = -4x/(12y) = -x/(3y)$ . At $(\sqrt{3}/2, 1/2)$ : $y' = -(\sqrt{3}/2)/(3/2) = -\sqrt{3}/3$ .
Ex. 54.27ModelingAnswer key
Use logarithmic differentiation to find $y'$ if $y = x^x$ ( $x > 0$ ).
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Let $y = x^x$ . Logarithmize: $\ln y = x \ln x$ . Differentiate: $y'/y = \ln x + x \cdot (1/x) = \ln x + 1$ . So $y' = y(\ln x + 1) = x^x(\ln x + 1)$ . The derivative of the "power-exponential" function involves both $\ln x$ (from the exponent) and $1$ (from the base).
Ex. 54.28ModelingAnswer key
Use logarithmic differentiation to find $y'$ if $y = x^{\sin x}$ ( $x > 0$ ). Evaluate at $x = \pi$ .
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Let $y = x^{\sin x}$ . Then $\ln y = \sin x \ln x$ . Differentiating: $y'/y = \cos x \ln x + \sin x / x$ . So $y' = x^{\sin x}(\cos x\,\ln x + \sin x/x)$ . Note that at $x = \pi$ : $y'(\pi) = \pi^0(\cos\pi \ln\pi + 0) = 1\cdot(-\ln\pi) = -\ln\pi$ .
Ex. 54.29Modeling
For $x^2 + y^2 = r^2$ , find $d^2y/dx^2$ in terms of $x$ , $y$ , and $r$ . Interpret the sign of $y''$ for $y > 0$ .
Solve online Active Calculus 2.0 · §2.7
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We already know $y' = -x/y$ . Differentiating again with respect to $x$ : $y'' = -(y - x\,y')/y^2$ . Substitute $y' = -x/y$ : $y'' = -(y - x(-x/y))/y^2 = -(y + x^2/y)/y^2 = -(y^2 + x^2)/y^3 = -r^2/y^3$ . For $y > 0$ (upper semicircle), $y'' < 0$ — the curve is concave down.
Ex. 54.30Modeling
For the ellipse $x^2/a^2 + y^2/b^2 = 1$ , calculate $dy/dx$ and $d^2y/dx^2$ .
Solve online OpenStax Calculus Volume 1 · §3.8
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For $x^2/a^2 + y^2/b^2 = 1$ , differentiating: $2x/a^2 + 2y\,y'/b^2 = 0$ , so $y' = -b^2 x/(a^2 y)$ . Differentiating again: $y'' = -b^2/a^2 \cdot (y - x\,y')/y^2$ . Substituting $y' = -b^2 x/(a^2 y)$ : $y'' = -b^2/a^2 \cdot (y + b^2 x^2/(a^2 y))/y^2 = -(b^2/a^2)(a^2 y^2 + b^2 x^2)/(a^2 y^3)$ . Since $a^2 y^2 + b^2 x^2 = a^2 b^2 (x^2/a^2 + y^2/b^2) = a^2 b^2$ for points on the ellipse: $y'' = -b^4/(a^2 y^3)$ .
Ex. 54.31Understanding
Why is the condition $F_y \neq 0$ necessary to apply the Implicit Function Theorem?
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Because the equation $F(x,y) = 0$ does not define a unique function of $x$ for the entire domainTo avoid division by zero in the formula $y' = -F_x/F_y$ Because $F_y = 0$ indicates a vertical tangent — the curve is not locally a function graphBecause $y$ must be a polynomial for implicit differentiation to work
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Ex. 54.32Understanding
What is the main advantage of implicit differentiation over isolating $y$ and differentiating explicitly?
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They are identical — implicit differentiation is just a different way of writing the same calculationImplicit differentiation works without isolating $y$ and unifies the curve's various branches into a single formulaExplicit differentiation never requires the chain ruleImplicit differentiation only works for polynomials
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Ex. 54.33Understanding
Use implicit differentiation to show that the tangent to the circle $x^2 + y^2 = r^2$ is always perpendicular to the radius at the point of tangency.
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Ex. 54.34UnderstandingAnswer key
For a curve $F(x,y)=0$ , explain the conditions under which the tangent line exists, possibly vertical, and when the point is singular.
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Ex. 54.35Understanding
Verify that differentiating $x^2 + y^2 = r^2$ implicitly gives the same result as differentiating $y = \pm\sqrt{r^2-x^2}$ explicitly.
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Ex. 54.36UnderstandingAnswer key
When implicitly differentiating $e^{xy} = x + y$ with respect to $x$ , what is $\frac{d}{dx}[e^y]$ ? Why is it not simply $e^y$ ?
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Ex. 54.37Challenge
For the curve $x^4 + y^4 = 1$ , find all points of horizontal and vertical tangency.
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For $x^4 + y^4 = 1$: differentiating, $4x^3 + 4y^3\,y' = 0$, so $y' = -x^3/y^3$. Horizontal tangent ($y'=0$): numerator $x^3=0 \Rightarrow x=0$, with $y = \pm 1$ on the curve. Vertical tangent (denominator $=0$): $y=0 \Rightarrow x = \pm 1$. Thus horizontal tangents at $(0, \pm 1)$ and vertical tangents at $(\pm 1, 0)$.
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Ex. 54.38Challenge
For the ellipse $x^2/a^2 + y^2/b^2 = 1$ , calculate $y''$ implicitly and simplify using the ellipse's equation. (Ans: $y'' = -b^4/(a^2 y^3)$ .)
Solve online OpenStax Calculus Volume 1 · §3.8
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For the ellipse $x^2/a^2 + y^2/b^2 = 1$: $y' = -b^2 x/(a^2 y)$. Differentiating again, using the quotient rule and substituting $y'$: $y'' = -b^2/a^2 \cdot (y - x\,y')/y^2$. Substituting $y'$: $y'' = -b^2/(a^2 y^2)(y + b^2 x^2/(a^2 y)) = -(b^2 a^2 y^2 + b^4 x^2)/(a^4 y^3)$. Using $x^2/a^2 + y^2/b^2=1$ implies $b^2 x^2/a^2 = b^2 - y^2$: $y'' = -b^4(x^2/a^2 + y^2/b^2)/(a^2 y^3) = -b^4/(a^2 y^3)$.
Ex. 54.39Challenge
For $\sin(xy) + \cos(x+y) = 1$ , calculate $dy/dx$ at $(0, 0)$ . Explain why the point is singular for the direct formula.
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For $\sin(xy) + \cos(x+y) = 1$, at $(0,0)$: check $\sin(0) + \cos(0) = 0 + 1 = 1$. Correct. Differentiating: $\cos(xy)(y + x\,y') - \sin(x+y)(1+y') = 0$. At $(0,0)$: $\cos(0)(0 + 0) - \sin(0)(1 + y') = 0 \Rightarrow 0 - 0 = 0$ — indeterminate. Expand around $(0,0)$ to second order: $xy - (x+y)^2/2 + \ldots \approx 0$. For $y = mx$: $mx^2 - x^2(1+m)^2/2 \approx 0$, so $m = (1+m)^2/2$. Solution by inspection: $m=0$ gives $0 = 1/2$... The point $(0,0)$ is singular for this system — $F_x = y\cos(xy) - \sin(x+y) = 0$ and $F_y = x\cos(xy) - \sin(x+y) = 0$ coincide at $(0,0)$. The directional derivative is $y'(0,0) = 0$ for the principal branch.
Ex. 54.40Proof
Proof. Prove that $(x^a)' = ax^{a-1}$ for arbitrary $a \in \mathbb{R}$ ( $x > 0$ ), using $x^a = e^{a\ln x}$ and the chain rule. Explain why the proof covers the case of irrational $a$ .
Solve online Active Calculus 2.0 · §2.7
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Write $x^a = e^{a \ln x}$ (definition of power with arbitrary real exponent, for $x > 0$ ). Differentiating: $(x^a)' = e^{a \ln x} \cdot \frac{d}{dx}[a \ln x] = x^a \cdot \frac{a}{x} = a\,x^{a-1}$ . This derivation holds for any $a \in \mathbb{R}$ , including irrational, which elementary limit-based proofs do not cleanly cover.
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1. Rewrite using the definition. For $x > 0$ and $a \in \mathbb{R}$ : $x^a = e^{a \ln x}$ . This is the definition of power with an arbitrary real exponent.
2. Apply the chain rule. $\dfrac{d}{dx}[e^{a \ln x}] = e^{a \ln x} \cdot \dfrac{d}{dx}[a \ln x] = e^{a \ln x} \cdot \dfrac{a}{x}$ .
3. Simplify. $e^{a \ln x} = x^a$ , so the result is $x^a \cdot a/x = a\,x^{a-1}$ .
4. Conclusion. $(x^a)' = a\,x^{a-1}$ for all $a \in \mathbb{R}$ and $x > 0$ . This is the power rule in its most general form.
Hint: The proof works because $e^u$ and $\ln x$ are well-defined functions for $x > 0$ . For $a \in \mathbb{Q}$ and $x > 0$ there are other proofs, but only this one covers the case $a \in \mathbb{R} \setminus \mathbb{Q}$ elementarily.

Sources

Active Calculus 2.0 — Boelkins · 2024 · §2.7 (Derivatives of Functions Given Implicitly). Primary source. CC-BY-NC-SA 4.0 license.
OpenStax Calculus Volume 1 — OpenStax · 2016 · §3.8 (Implicit Differentiation). CC-BY-NC-SA 4.0 license.
APEX Calculus — Hartman et al. · 2024 · v5 · §2.6 (Implicit Differentiation). CC-BY-NC 4.0 license.