3.10 Implicit Differentiation

We have developed the basic techniques for calculating a derivative \(\frac{dy}{dx}\) when \(y\) is given in terms of \(x\) by a formula—such as \(y = x^{3} + 1\). But suppose that \(y\) is determined instead by an equation such as

\[y^4 + xy = x^3 - x +2\tag{1}\]

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In this case, we say that \(y\) is defined implicitly as a function of \(x\), as opposed to explicitly as \(y=f(x)\). How can we find the slope of the tangent line at a point on the graph (Figure 3.59)? Although it may be difficult or even impossible to solve for \(y\) explicitly as a function of \(x\), we can find \(\frac{dy}{dx}\) using the method of implicit differentiation.

Figure 3.59: Plot of \(y^{4} + xy = x^{3} - x + 2\)

To illustrate, consider the equation of the unit circle (Figure 3.60):

\[x^{2} + y^{2} = 1\]

Figure 3.60: The tangent line to the unit circle \(x^2+y^2=1\) at \(P(\frac{3}{5},\frac{4}{5})\) has slope \(-\frac{3}{4}\).

Compute \(\frac{dy}{dx}\) by taking the derivative of both sides of the equation:

\begin{align*} \frac{d}{dx}(x^2+y^2)&=\frac{d}{dx}(1)\\ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2)&=0\\ 2x + \frac{d}{dx}(y^2)&=0 \tag{2} \end{align*}

How do we handle the term \(\frac{d}{dx}(y^2)\)? We use the Chain Rule. Think of \(y\) as a function \(y = f(x)\). Then \(y^{2} = f(x)^{2}\) and by the Chain Rule,

\[ \frac{d}{dx}y^2 = \frac{d}{dx}f(x)^2 = 2f(x)\frac{df}{dx} = 2y\frac{dy}{dx} \]

or, in Leibniz notation,

\[ \frac{d}{dx}y^2 = \frac{d}{dy}y^2\frac{dy}{x} = 2y\frac{dy}{dx} \]

Equation (2) becomes \(\displaystyle 2x+2y\frac{dy}{dx} = 0\), and we can solve for \(\displaystyle \frac{dy}{dx}\) if \(y \neq 0\):

\[ \boxed{\dfrac{dy}{dx} = -\dfrac{x}{y}}\tag{3} \]

EXAMPLE 1

Use Eq. (3) to find the slope of the tangent line at the point \(P (\frac{3}{5},\frac{4}{5})\) on the unit circle.

Solution Set \(\displaystyle x= \frac{3}{5}\) and \(\displaystyle y=\frac{4}{5}\) in Eq. (3):

\[ \left.\frac{dy}{dx}\right|_P = -\dfrac{x}{y} = -\dfrac{\frac{3}{5}}{\frac{4}{5}} = -\frac{3}{4} \]

In this particular example, we could have computed dy/dx directly, without implicit differentiation. The upper semicircle is the graph of and

This formula expresses \(\frac{dy}{dx}\) in terms of \(x\) alone, whereas Eq. (3) expresses \(\frac{dy}{dx}\) in terms of both \(x\) and \(y\), as is typical when we use implicit differentiation. The two formulas agree because \(y=\sqrt{1-x^2}\).

Notice what happens if we insist on applying the Chain Rule to \(\frac{d}{dy}\sin y\). The extra factor appears, but it is equal to \(1\):

\[ \frac{d}{dy}\sin y = (\cos y)\frac{dy}{dy} = \cos y \]

Before presenting additional examples, let’s examine again how the factor \(\frac{dy}{dx}\) arises when we differentiate an expression involving \(y\) with respect to \(x\). It would not appear if we were differentiating with respect to \(y\). Thus,

\begin{align*} \frac{d}{dy}\sin y &=\cos y &\text{but}& &\frac{d}{dx}\sin y &= (\cos y)\frac{dy}{dx}\\ \frac{d}{dy}y^4 &= 4y^3 &\text{but} && \frac{d}{dx}y^4 &= 4y^3\frac{dy}{dx} \end{align*}

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Similarly, the Product Rule applied to \(xy\) yields

\[ \frac{d}{dx}(xy) = x\frac{dy}{dx} + y\frac{dx}{dx} = x \frac{dy}{dx}+y \]

The Quotient Rule applied to \(\displaystyle \frac{t^{2}}{y}\) yields

\[ \frac{d}{dt}\left(\frac{y\frac{d}{dt}t^2 - t^2 \frac{dy}{dt}}{y^2}\right) = \frac{2ty - t^2\frac{dy}{dt}}{y^2} \]

EXAMPLE 2

Find an equation of the tangent line at the point \(P = (1, 1)\) on the curve (Figure 3.59)

\[ y^{4} + xy = x^{3} - x + 2 \]

Solution We break up the calculation into two steps.

Step 1. Differentiate both sides of the equation with respect to x.

\begin{align*} \frac{d}{dx}y^{4} + \frac{d}{dx}(xy) &= \frac{d}{dx}(x^{3} - x + 2)\\[5pt] 4y^3\frac{dy}{dx} + \left(x\frac{dy}{dx} + y\right)&=3x^2 -1\tag{4} \end{align*}

Step 2. Solve for \(\frac{dy}{dx}\).

Move the terms involving \(\frac{dy}{dx}\) in Eq. (4) to the left and place the remaining terms on the right:

\[ 4y^3 \frac{dy}{dx} + x\frac{dy}{dx} = 3x^2-1 - y \]

Then factor out \(\frac{dy}{dx}\) and divide:

\begin{align*} \left(4y^3 + x\right)\frac{dy}{dx}&=3x^2-1 - y\\ \frac{dy}{dx}&= \frac{3x^2-1 - y}{4y^3 + x} \end{align*}

To find the derivative at \(P = (1, 1)\), apply Eq. (5) with \(x = 1\) and \(y = 1\):

\[ \left.\frac{dy}{dx}\right|_{(1,1)} = \frac{3\cdot1^2-1 - 1}{4\cdot1^3 + 1}=\frac{1}{5} \]

An equation of the tangent line is \(y-1 = \frac{1}{5}(x-1)\) or \(y =\frac{1}{5}x + \frac{4}{5}\).

CONCEPTUAL INSIGHT

The graph of an equation does not always define a function because there may be more than one \(y\)-value for a given value of \(x\). Implicit differentiation works because the graph is generally made up of several pieces called branches, each of which does define a function (a proof of this fact relies on the Implicit Function Theorem from advanced calculus). For example, the branches of the unit circle \(x^{2} + y^{2} = 1\) are the graphs of the functions \(y=-\sqrt{1-x^2}\) and \(y=\sqrt{1-x^2}\). Similarly, the graph in Figure 3.61 has an upper and a lower branch. In most examples, the branches are differentiable except at certain exceptional points where the tangent line may be vertical.

Figure 3.61: Each branch of the plot of \(y^{4} + xy = x^{3} - x + 2\) defines a function of \(x\).

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EXAMPLE 3

Find the slope of the tangent line at the point \(P = (1, 1)\) on the graph of \(e^{x - y} = 2x^{2} - y^{2}\).

Solution We follow the steps of the previous example, this time writing \(y'\) for \(\frac{dy}{dx}\):

\begin{align*} \frac{d}{dx}e^{x - y} &= \frac{d}{dx}(2x^{2} - y^{2})\\[5pt] e^{x-y}(1-y')& = 4x - 2yy'&&(\text{Chain Rule applied to }e^{x-y})\\[5pt] e^{x-y} - e^{x-y}y' &= 4x - 2yy'\\[5pt] (2y - e^{x-y})y'&= 4x - e^{x-y} &&(\text{Place all }y'\text{-terms on left})\\[5pt] y' &= \frac{4x - e^{x-y}}{2y - e^{x-y}} \end{align*}

The slope of the tangent line at \(P = (1, 1)\) is (Figure 3.62)

\[ \left.\frac{dy}{dx}\right|_(1,1) = \frac{4(1) - e^{1-1}}{2(1) - e^{1-1}}=\frac{4-1}{2-1} = 3 \]

Figure 3.62: Plot of \(e^{x-y} = 2x^{2} - y^{2}\).

Question 3.25 Implicit Differentiation Progress Check 1

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3
Correct.
Try again. Differentiate implicitly, assuming \(y\) is a function of \(x\). Thus don't forget to use the chain rule when differentiating \(y^3\).
Incorrect.

EXAMPLE 4 Shortcut to Derivative at a Specific Point

Calculate \(\left.\frac{dy}{dt}\right|_P\) at the point \(P=(0,\frac{5\pi}{2})\) on the curve (Figure 3.63):

\[ y \ cos(y + t + t^{2}) = t^{3} \]

Solution As before, differentiate both sides of the equation (we write \(y'\) for \(\frac{dy}{dx}\)):

\begin{align*} \frac{d}{dt}y\cos(y + t + t^{2})&= \frac{d}{dt}t^{3}\\[5pt] y'\cos(y + t + t^{2}) - y\sin(y + t + t^{2})(y'+1+2t) &=3t^2\tag{6} \end{align*}

We could continue to solve for \(y'\), but that is not necessary. Instead, we can substitute \(t = 0\), \(y=\frac{5\pi}{2}\) directly in Eq. (6) to obtain

\begin{align*} y'\cos\left(\frac{5\pi}{2} + 0 + 0^{2}\right) - \frac{5\pi}{2}\sin\left(\frac{5\pi}{2} + 0 + 0^{2}\right)(y'+1+0) &=0\\[5pt] 0 - \left(\frac{5\pi}{2}\right)(1)(y' + 1) = 0 \end{align*}

This gives us \(y' + 1 = 0\) or \(y' = -1\).

Figure 3.63: Plot of \( y \cos(y + t + t^{2}) = t^{3} \). The tangent line at \( P(0,\frac{5\pi}{2})\) has slope -1. move the slider to see the tangent lines to the curve at other points.

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3.10.1 Section 3.10 Summary

Step 1. Take the derivative of both sides of the equation with respect to \(x\).

Step 2. Solve for \(\frac{dy}{dx}\) by collecting the terms involving \(\frac{dy}{dx}\) on one side and the remaining terms on the other side of the equation.

\[ \frac{d}{dx}\sin y = (\cos y)\frac{dy}{dx} \]