OBJECTIVES

When you finish this section, you should be able to:

1. Find a derivative using implicit differentiation (p. 209)
2. Find higher-order derivatives using implicit differentiation (p. 212)
3. Differentiate functions with rational exponents (p. 213)
4. Find the derivative of an inverse function (p. 214)
5. Differentiate the inverse trigonometric functions (p. 216)

NEED TO REVIEW?

The implicit form of a function is discussed in Section P.1, p. 4.

Find $\dfrac{dy}{dx}$ if $xy-4=0$.

Solution (a) To differentiate implicitly, we assume $y$ is a differentiable function of $x$ and differentiate both sides with respect to $x$. $$\begin{eqnarray*} \begin{array}{rl@{\qquad}l} \dfrac{d}{dx}(xy-4) &=\dfrac{d}{dx}0 & {\color{#0066A7}{\hbox{Differentiate both sides with respect to}\ {x}.}}\\ \dfrac{d}{dx}(xy) -\dfrac{d}{dx}4 &= 0 & {\color{#0066A7}{\hbox{Use the Difference Rule.}}}\\ x\cdot \dfrac{d}{dx}y+\left(\dfrac{d}{dx}x\right) y-0 &= 0 & {\color{#0066A7}{\hbox{Use the Product Rule.}}}\\ x\dfrac{dy}{dx}+y & =0 & {\color{#0066A7}{\hbox{Simplify}.}} \\ \dfrac{dy}{dx} & = -\dfrac{y}{x} & {\color{#0066A7}{\hbox{Solve for}\, \dfrac{dy}{dx}.}} & (1) \end{array} \end{eqnarray*}$$

(b) Solve $xy-4=0$ for $y$, obtaining $y=\dfrac{4}{x}=4x^{-1}$. Then $$\begin{equation*} \dfrac{dy}{dx}=\dfrac{d}{dx}(4x^{-1}) =-4x^{-2}=-\dfrac{4}{x^{2}}\tag{2} \end{equation*}$$

(c) At first glance, the results in (1) and (2) appear to be different. However, since $xy-4=0,$ or equivalently, $y=\dfrac{4}{x},$ the result from (1) becomes $$\begin{eqnarray*} && \dfrac{dy}{dx}\underset{\underset{\color{#0066A7}{(1)}}{\color{#0066A7}{\uparrow}}}{=}-\dfrac{y}{x}\underset{\underset{\color{#0066A7}{y=\tfrac{4}{x}}}{\color{#0066A7}{\uparrow}}}{=} -{\dfrac{\dfrac{4}{x}}{x}}=-\dfrac{4}{x^{2}} \end{eqnarray*}$$ which is the same as (2).

NOW WORK

Problem 17.

To differentiate an implicit function:

• Assume that $y$ is a differentiable function of $x$.
• Differentiate both sides of the equation with respect to $x$.
• Solve the resulting equation for $y^\prime =\dfrac{dy}{dx}.$

Find $\dfrac{dy}{dx}$ if $3x^{2}+4y^{2}=2x$.

Solution We assume that $y$ is a differentiable function of $x$ and differentiate both sides with respect to $x$. $$\begin{eqnarray*} \begin{array}{rl@{\qquad}l} \dfrac{d}{dx} ( 3x^{2}+4y^{2}) &= \dfrac{d}{dx}(2x) & {\color{#0066A7}{\hbox{Differentiate both sides with respect to}\; x.}}\\ \dfrac{d}{dx} (3x^{2}) +\dfrac{d}{dx}( 4y^{2}) &= 2 & {\color{#0066A7}{\hbox{Sum Rule}.}} \\ 3\dfrac{d}{dx}x^{2}+4\dfrac{d}{dx}y^{2} &=2 &{\color{#0066A7}{\hbox{Constant Multiple Rule}.}}\\ 6x+4\!\left( 2y\dfrac{dy}{dx}\right) &= 2 &{\color{#0066A7}{\dfrac{d}{dx}{y}^{{ 2}}{=2y} \dfrac{dy}{dx}}}\\ 6x+8y\dfrac{dy}{dx} &= {2} & {\color{#0066A7}{\hbox{Simplify.}}}\\ \dfrac{dy}{dx} &= \dfrac{2-6x}{8y}=\dfrac{1-3x}{4y} & {\color{#0066A7}{\hbox{Solve for}\; \dfrac{dy}{dx}.}} \end{array} \end{eqnarray*}$$ provided $y\ne 0$.

NOW WORK

Problem 15.

Find an equation of the tangent line to the graph of the ellipse $3x^{2}+4y^{2}=2x$ at the point $\left( \dfrac{1}{2},-\dfrac{1}{4}\right)$.

Solution First we find the slope of the tangent line. We use the result from Example 2, and evaluate $\dfrac{dy}{dx}=\dfrac{1-3x}{4y}$ at $\left( \dfrac{1}{2},-\dfrac{1}{4}\right)$. $$\begin{eqnarray*} && \dfrac{dy}{dx}=\dfrac{1-3x}{4y} \underset{\underset{\color{#0066A7}{x=\tfrac{1}{2},y=-\dfrac{1}{4}}}{\color{#0066A7}{{\left\uparrow{\vphantom{\vrule width0pc height12.5pt depth0pt}}\right.}}}}{=} \dfrac{1-3\cdot \dfrac{1}{2}}{4\,{\cdot}\,\!\left(-\dfrac{1}{4}\right) }=\dfrac{1}{2}\\ \end{eqnarray*}$$

The slope of the tangent line to the graph of $3x^{2}+4y^{2}=2x$ at the point $\left( \dfrac{1}{2},-\dfrac{1}{4}\right)$ is $\dfrac{1}{2}$. An equation of the tangent line is $$\begin{eqnarray*} y+\dfrac{1}{4} &=&\dfrac{1}{2}\!\left( x-\dfrac{1}{2}\right) \\ y &=&\dfrac{1}{2}x-\dfrac{1}{2} \end{eqnarray*}$$

NOW WORK

Problem 63.

Solution (a) We use implicit differentiation: $$\begin{eqnarray*} \begin{array}{rl@{\qquad}l} \dfrac{d}{dx}( e^{y}\cos x) &= \dfrac{d}{dx}(x+1) \\ e^{y}\cdot \dfrac{d}{dx}( \cos x) +\left( \dfrac{d}{dx} e^{y}\right) \cdot \cos x &= 1 & {\color{#0066A7}{\hbox{Use the Product Rule.}}} \\ e^{y}( -\sin x) +e^{y}y^\prime \cdot \cos x &= 1 \\ ( e^{y}\cos x) y^\prime &= 1+e^{y}\sin x \\ y^\prime &= \dfrac{1+e^{y}\sin x}{e^{y}\cos x} \end{array} \end{eqnarray*}$$

(b) At the point $(0,0)$, the derivative $y^\prime$ is $\dfrac{1+e^{0}\sin 0}{e^{0}\cos 0}=\dfrac{1+1\cdot 0}{1\cdot 1}=1$, so the slope of the tangent line to the graph at $(0,0)$ is $1$. An equation of the tangent line to the graph at the point $(0,0)$ is $y=x.$

Use implicit differentiation to find $y^\prime$ and $y^{\prime\prime}$ if $y^{2}-x^{2}=5$. Express $y^{\prime\prime}$ in terms of $x$ and $y$.

Solution First, we assume there is a differentiable function $y=f( x)$ that satisfies $y^{2}-x^{2}=5$. Now we find $y^\prime$.



provided $y≠ 0$.

Equations (3) and (4) both involve $y^\prime$. Either one can be used to find $y^{\prime\prime} .$ We use (3) because it avoids differentiating a quotient. $$\begin{eqnarray*} \dfrac{d}{dx} ( 2yy^\prime -2x) &=&\dfrac{d}{dx}0 \nonumber\\ \dfrac{d}{dx} ( yy^\prime ) -\dfrac{d}{dx}x &=&0 \nonumber\\ y\cdot \dfrac{d}{dx}y^\prime +\left( \dfrac{d}{dx}y\right) y^\prime -1 &=&0 \nonumber\\ yy^{\prime\prime} +( y^\prime ) ^{2}-1 &=&0 \nonumber\\ y^{\prime\prime} &=&\dfrac{1-( y^\prime ) ^{2}}{y}\tag{5} \end{eqnarray*}$$ provided $y≠ 0$. To express $y^{\prime\prime}$ in terms of $x$ and $y,$ we use (4) and substitute for $y^\prime$ in (5).

NOW WORK

Problem 59.

THEOREM Power Rule for Rational Exponents

If $y=x^{p/q}$, where $\dfrac{p}{q}$ is a rational number, then $$\begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{y^\prime =\dfrac{d}{dx}x^{p/q}=\dfrac{p}{q}\cdot x^{\left(p/q\right) -1}}\tag{6} \end{equation*}$$ provided that $x^{p/q}$ and $x^{(p/q) -1}$ are defined.

Proof

We begin with the function $y=x^{p/q},$ where $p$ and $q>0$ are integers. Now, we raise both sides of the equation to the power $q$ to obtain $$y^{q}=x^{p}$$

This is the implicit form of the function $y=x^{p/q}.$ Assuming $y$ is differentiable*, we can differentiate implicitly, obtaining $$\begin{eqnarray*} \frac{d}{dx}y^q &=& \frac{d}{dx} x^p\\ qy^{q-1}y^\prime &=& px^{p-1} \end{eqnarray*}$$

Now solve for $y^\prime$. $$\begin{eqnarray*} && y^\prime =\dfrac{px^{p-1}}{qy^{q-1}}\underset{\underset{\color{#0066A7}{y=x^{p/q}}}{\color{#0066A7}{\uparrow}}}{=}\dfrac{p}{q}\cdot \dfrac{x^{p-1}}{( x^{p/q}) ^{q-1}}=\dfrac{p}{q} \cdot \dfrac{x^{p-1}}{x^{p-(p/q) }}=\dfrac{p}{q}\cdot x^{p-1-[ p-(p/q)] }=\dfrac{p}{q}\cdot x^{(p/q)-1}\\ \end{eqnarray*}$$

NOW WORK

Problem 31.

THEOREM Power Rule for Functions

If $u$ is a differentiable function of $x$ and $r$ is a rational number, then $$\bbox[5px, border:1px solid black, #F9F7ED] {\dfrac{d}{dx}[u(x)]^{r}=r[u(x)] ^{r-1}u^\prime (x)}$$ provided $u^{r}$ and $u^{r-1}$ are defined.

NOW WORK

Problem 39.

NEED TO REVIEW?

Inverse functions are discussed in Section P.4, pp. 32-37.

THEOREM Derivative of an Inverse Function

Let $y=f( x)$ and $x=g( y)$ be inverse functions. Suppose $f$ is differentiable on an open interval containing $x_{0}$ and $y_{0}=f( x_{0}) .$ If $f^\prime ( x_{0}) ≠ 0$, then $g$ is differentiable at $y_{0}=f( x_{0})$ and $$\begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{ \dfrac{d}{dy}g(y_{0}) =g^\prime (y_{0}) =\dfrac{1}{f^\prime (x_{0})}}\tag{7} \end{equation*}$$ where the notation $\dfrac{d}{dy}g( y_{0})=g^\prime(y_{0})$ means the value of $\dfrac{d}{dy}g( y)$ at $y_{0},$ and the notation $f^\prime ( x_{0})$ means the value of $f^\prime ( x)$ at $x_{0}$.

In Leibniz notation, formula (7) has the simple form $$\bbox[5px, border:1px solid black, #F9F7ED]{ \dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}}$$

• If $y_{0}=f( x_{0})$ and $f^\prime ( x_{0}) ≠ 0$ exists, then $\dfrac{d}{dy}g( y)$ exists at $y_{0}.$
• It gives formula $( 7)$ for finding $\dfrac{d}{dy}g(y)$ at $y_{0}$ without knowing a formula for $g=f^{-1}$, provided we can find $x_{0}$ and $f^\prime ( x_{0}) .$

The function $f( x) =x^{5}+x$ has an inverse function $g$. Find $g^\prime (2)$.

Solution Using (7) with $y_{0}=2$, we get $$g^\prime (2) =\dfrac{1}{f^\prime (x_{0})}\qquad \hbox{where}\ 2=f( x_{0})$$

A solution of the equation $$f( x_{0}) =x_{0}^{5}+x_{0}=2$$ is $x_{0}=1$. Since $f^\prime ( x) =5x^{4}+1$, then $f^\prime (x_{0})= f^\prime (1) =6$ and $$g^\prime (2) =\dfrac{1}{f^\prime (1) }=\dfrac{1}{6}$$

NOW WORK

Problem 71.

NEED TO REVIEW?

Inverse trigonometric functions are discussed in Section P.7, pp. 58-61.

THEOREM Derivative of $y=\sin ^{-1}x$

The derivative of the inverse sine function $y=\sin ^{-1}x$ is $$\bbox[5px, border:1px solid black, #F9F7ED]{ y^\prime = \dfrac{d}{dx}\sin ^{-1}x=\dfrac{1}{\sqrt{1-x^{2}}} \qquad -1< x< 1}$$

Find $y^\prime$ if:
(a) $y=\sin ^{-1}(4x^{2})$ $\quad$ (b) $y=e^{\sin ^{-1}x}$

Solution (a) If $y=\sin ^{-1}u$ and $u=4x^{2}$, then $\dfrac{dy}{du}=\dfrac{1}{\sqrt{1-u^{2}}}$ and $\dfrac{du}{dx}=8x.$ By the Chain Rule, $$\begin{eqnarray*} && y^\prime =\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=\left( \dfrac{1}{ \sqrt{1-u^{2}}}\right) (8x) = \dfrac{8x}{\sqrt{1-16x^{4}}}\qquad {\color{#0066A7}{\hbox{\(u=4x^{2}\)}}} \end{eqnarray*}$$

(b) If $y=e^{u}$ and $u=\sin ^{-1}x$, then $\dfrac{dy}{du}=e^{u}$ and $\dfrac{du}{dx}=\dfrac{1}{\sqrt{1-x^{2}}}$. By the Chain Rule, $$\begin{eqnarray*} && y^\prime =\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=e^{u}\cdot \dfrac{ 1}{\sqrt{1-x^{2}}} \underset{\underset{\color{#0066A7}{\hbox{\(u=\sin ^{-1}x\)}}}{\color{#0066A7}{\uparrow}}}{=} \dfrac{ e^{\sin ^{-1}x}}{\sqrt{1-x^{2}}} \\ \end{eqnarray*}$$

NOW WORK

Problem 51.

THEOREM Derivative of ${y=\tan ^{-1}x}$

The derivative of the inverse tangent function $y=\tan ^{-1}x$ is $$\bbox[5px, border:1px solid black, #F9F7ED]{ y^\prime =\dfrac{d}{dx}\tan ^{-1}x=\dfrac{1}{1+x^{2}}}$$

The domain of $y^\prime$ is all real numbers.

Find $y^\prime$ if:
(a) $y=\tan ^{-1}( 4x)$ $\quad$ (b) $y=\sin ( \tan ^{-1}x)$

Solution (a) Let $y=\tan ^{-1}u$ and $u=4x.$ Then $\dfrac{dy}{du}=\dfrac{1}{1+u^{2}}$ and $\dfrac{du}{dx}=4.$ By the Chain Rule, $$y^\prime =\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=\dfrac{1}{1+u^{2}} \cdot 4=\dfrac{4}{1+16x^{2}}$$

(b) Let $y=\sin u$ and $u=\tan ^{-1}x$. Then $\dfrac{dy}{du}=\cos u$ and $\dfrac{du}{dx}=\dfrac{1}{1+x^{2}}$. By the Chain Rule, $$y^\prime =\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}=\cos u\cdot \dfrac{1}{1+x^{2}}=\dfrac{\cos ( \tan ^{-1}x) }{1+x^{2}}$$

NOW WORK

Problem 55.

THEOREM Derivative of ${y=\sec ^{-1}x}$

The derivative of the inverse secant function $y=\sec ^{-1}x$ is $$\bbox[5px, border:1px solid black, #F9F7ED]{ y^\prime =\dfrac{d}{dx}\sec ^{-1}x=\dfrac{1}{x\sqrt{x^{2}-1}}\qquad \vert x\vert > 1 }$$

• Since $\cos ^{-1}x=\dfrac{\pi }{2}-\sin ^{-1}x$, then $\dfrac{d}{dx} \cos ^{-1}x=-\dfrac{1}{\sqrt{1-x^{2}}}$, where $\vert x\vert\,{< }\,1$.
• Since $\cot ^{-1}x=\dfrac{\pi }{2}-\tan ^{-1}x,$ then $\dfrac{d}{dx} \cot ^{-1}x=-\dfrac{1}{1+x^{2}}$, where $-\infty\,{< }\,x\,{< }\,\infty$.
• Since $\csc ^{-1}x=\dfrac{\pi }{2}-\sec ^{-1}x,$ then $\dfrac{d}{dx} \csc ^{-1}x=-\dfrac{1}{x\sqrt{x^{2}-1}}$, where $\vert x\vert\,{>}\,1$.