Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 4

Differentiation

The Chain Rule

The chain rule is applied in the differentiation of the composition of two or more functions. We will start by identifying the composition of a power function $f\left(u\right)=u^r$ and a function $u=g\left(x\right)$. That is,

$f\circ g\left(x\right)=f\left(g\left(x\right)\right)={\left(g\left(x\right)\right)}^r\text{.}$

It is important that you recognize a function of this form as a composition of these two functions.

Example 4.39. The function $h\left(x\right)=\sqrt{4x-6}$ is the composition of

$f\left(u\right)=u^{1∕2}\text{ and }u=g\left(x\right)=4x-6\text{.}$

To see this, note that

$h\left(x\right)=f\circ g\left(x\right)=f\left(g\left(x\right)\right)={\left(4x-6\right)}^{1∕2}\text{.}$

Example 4.40. The function $h\left(x\right)={\tan }^5\left(3x\right)$ is the composition of

$f\left(u\right)=u^5\text{ and }u=g\left(x\right)=\tan \left(3x\right)\text{,}$

since

$h\left(x\right)=f\left(g\left(x\right)\right)={\left(\tan \left(3x\right)\right)}^5\text{.}$

The composition of a trigonometric function (as outside function) and a function $g\left(x\right)$ (as an inside function) has the form $\text{TrigFctn}\left(g\left(x\right)\right)\text{.}$

Example 4.41. The function $\cos \left(x^2-5\right)$ is the composition of $\mathrm{cos }t$ and $g\left(x\right)=x^2-5$; that is, $\cos \left(g\left(x\right)\right)\text{.}$

Example 4.42. The function $\sec \left(\mathrm{cot }x\right)$ is the composition of $\mathrm{sec }t$ and $g\left(x\right)=\mathrm{cot }x$, that is, $\sec \left(g\left(x\right)\right)$.

Example 4.43. The function

${\displaystyle R\left(x\right)=\frac{1}{{\sec }^2\left(\sqrt{x}\right)}}$

has the form

${\displaystyle R\left(x\right)={\left(\sec \left(\sqrt{x}\right)\right)}^{-2}\text{;}}$

hence, it is the composition of the power function $f\left(u\right)=u^{-2}$ and the function $u=g_1\left(x\right)=\sec \left(\sqrt{x}\right)$.

At the same time, the function $g_1\left(x\right)$ is the composition of the functions $\mathrm{sec }t$ and $g_2\left(x\right)=\sqrt{x}$.

Hence,

$R\left(x\right)=f\left(\sec \left(g_2\left(x\right)\right)\right)={\left[\sec \left(g\left(x\right)\right)\right]}^{-2}\text{.}$

Our problem now is to find the derivative of such functions. The rule that we apply is the chain rule.

The Chain Rule:   ${\displaystyle \frac{d}{dx}f\circ g\left(x\right)=f^{\prime }\left(g\left(x\right)\right)g^{\prime }\left(x\right)}$

On page 152 of the textbook, the author makes an attempt to explain why the chain rule works. To apply this rule, we must understand it. The term $f^{\prime }\left(g\left(x\right)\right)$ is the composition of the derivative function $f^{\prime }$ (outside function) and the function $g\left(x\right)$ (inside function); that is $f^{\prime }\left(g\left(x\right)\right)=f^{\prime }\circ g\left(x\right)$. So, we must obtain the derivative of the outside function, and then make a composition of this derivative function (as an outside function) with the function $g\left(x\right)\text{.}$

Example 4.44. The function $h\left(x\right)=\sqrt{4x-6}$ (see Example 4.39, above) is the composition of $f\left(u\right)=u^{1∕2}$ and $g\left(x\right)=4x-6$.

We have

${\displaystyle f^{\prime }\left(u\right)=\frac{1}{2\sqrt{u}}\text{,}}$

so

${\displaystyle f^{\prime }\left(g\left(x\right)\right)=\frac{1}{2\sqrt{g\left(x\right)}}=\frac{1}{2\sqrt{4x-6}}\text{,}}$

and since $g^{\prime }\left(x\right)=4$, the derivative is

${\displaystyle h^{\prime }\left(x\right)=\left(\frac{1}{2\sqrt{4x-6}}\right)\left(4\right)=\frac{2}{\sqrt{4x-6}}\text{.}}$

Example 4.45. The function $h\left(x\right)={\sin }^{3}x$ is the composition of $f\left(u\right)=u^3$ and $g\left(x\right)=\mathrm{sin }x$.

Since $f^{\prime }\left(u\right)=3u^2$, then $f^{\prime }\left(g\left(x\right)\right)=3{\left(g\left(x\right)\right)}^2=3{\mathrm{ sin}}^{\mathrm{2 }}x$, and $g^{\prime }\left(x\right)=\mathrm{cos }x$. So, $h^{\prime }\left(x\right)=3{\mathrm{ sin}}^{\mathrm{2 }}x cos x\text{.}$ See Example 4.38.

A function of the form $h\left(x\right)={\left(g\left(x\right)\right)}^r$ is the composition of $f\left(u\right)=u^r$ and $u=g\left(x\right)$. Since $f^{\prime }\left(u\right)=r{u}^{r-1}$, we have $f^{\prime }\left(g\left(x\right)\right)=r{\left(g\left(x\right)\right)}^{r-1}$, and then

${\displaystyle \frac{d}{dx}{\left(g\left(x\right)\right)}^r=r{\left(g\left(x\right)\right)}^{r-1}{g}^{\prime }\left(x\right)\text{.}}$

This is known as the general power rule.

The General Power Rule:   ${\displaystyle \frac{d}{dx}{(g(x))}^r=r{(g(x))}^{r-1}{g}^{\prime }(x)\text{.}}$

Warning: When applying the general power rule, do not forget to multiply $r{(g(x))^{r - 1}}$ by the derivative $g'(x)$. Observe that, to apply the general power rule, we must identify the function $g(x)$ and the power $r$.

Example 4.46. The function $f\left(x\right)=\sqrt{{\cos }^{\mathrm{3 }}x+5}$ is equal to $f\left(x\right)={\left({\cos }^{\mathrm{3 }}x+5\right)}^{1∕2}$. Hence, the power is $r=1∕2$ and $g\left(x\right)={\cos }^{\mathrm{3 }}x+5$.

We will find the derivative of $g$ first. Alas, we must use the general power rule, since $g\left(x\right)={\left(\mathrm{cos }x\right)}^3+5$.

Hence, $g^{\prime }\left(x\right)=-3{\left(\mathrm{cos }x\right)}^2\mathrm{ sin }x\text{.}$ [Agree?]

Then,

${\displaystyle \begin{array}{ll}\hfill f^{\prime }\left(x\right)& ={\displaystyle \frac{1}{2}}{\left({\cos }^{3}x+5\right)}^{-1/2}\left(-3{\left(\cos x\right)}^2\sin x\right)\hfill \\ \hfill & =-{\displaystyle \frac{3}{2}}{\displaystyle \frac{{\cos }^{2}x\sin x}{\sqrt{{\cos }^{3}x+5}}}\text{.}\hfill \end{array}}$

Example 4.47. For the function $R\left(x\right)={\left(\sec \left(\sqrt{x}\right)\right)}^{-2}$ in Example 4.42, we have

${\displaystyle \frac{d}{dt}\mathrm{ sec }t=\mathrm{sec }t\mathrm{ tan }t\text{ and }\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\text{,}}$

and by the chain rule, we conclude that

${\displaystyle \frac{d}{dx}\mathrm{ sec}\left(\sqrt{x}\right)=\sec \left(\sqrt{x}\right) tan\left(\sqrt{x}\right)\left(\frac{1}{2\sqrt{x}}\right)\text{.}}$

Hence, by the general power rule,

${\displaystyle \begin{array}{ll}\hfill R^{\prime }\left(x\right)& =-2(\sec {\left(\sqrt{x}\right)}^{-3}\left[\sec \left(\sqrt{x}\right)\tan \left(\sqrt{x}\right)\left({\displaystyle \frac{1}{2\sqrt{x}}}\right)\right]\hfill \\ \hfill & ={\displaystyle \frac{-\tan \left(\sqrt{x}\right)}{{\sec }^2\left(\sqrt{x}\right)\sqrt{x}}}\text{.}\hfill \end{array}}$

Some particular cases of the general power rule are worth keeping in mind, because we make frequent use of them in this course.

${\displaystyle \begin{array}{ll}\hfill \text{If }f\left(x\right)& ={\displaystyle \frac{1}{g\left(x\right)}}\text{, then }r=-1\text{ and}\hfill \\ \hfill & {\displaystyle \frac{d}{dx}}{\displaystyle \frac{1}{g\left(x\right)}}={\displaystyle \frac{d}{dx}}{\left(g\left(x\right)\right)}^{-1}=-{\displaystyle \frac{g^{\prime }\left(x\right)}{{\left(g\left(x\right)\right)}^2}}\text{.}\hfill \end{array}}$ (4.1)

${\displaystyle \begin{array}{ll}\hfill \text{If }f\left(x\right)& =\sqrt{g\left(x\right)}\text{, then }r={\displaystyle \frac{1}{2}}\text{ and}\hfill \\ \hfill & {\displaystyle \frac{d}{dx}}\sqrt{g\left(x\right)}={\displaystyle \frac{d}{dx}}{\left(g\left(x\right)\right)}^{1/2}={\displaystyle \frac{g^{\prime }\left(x\right)}{2\sqrt{g\left(x\right)}}}\text{.}\hfill \end{array}}$ (4.2)

${\displaystyle \begin{array}{ll}\hfill \text{If }f\left(x\right)& ={\displaystyle \frac{1}{\sqrt{g\left(x\right)}}}\text{, then }r=-{\displaystyle \frac{1}{2}}\text{ and}\hfill \\ \hfill & {\displaystyle \frac{d}{dx}}{\displaystyle \frac{1}{\sqrt{g\left(x\right)}}}={\displaystyle \frac{d}{dx}}{\left(g\left(x\right)\right)}^{-1/2}=-{\displaystyle \frac{g^{\prime }\left(x\right)}{2{\left(g\left(x\right)\right)}^{3/2}}}\text{.}\hfill \end{array}}$ (4.3)

Example 4.48. In the function ${\displaystyle f\left(x\right)=\frac{\sqrt{3}}{x^4\mathrm{ tan }x}\text{,}}$ the derivative of $g\left(x\right)=x^4\mathrm{ tan }x$ is obtained by applying the product rule

$g^{\prime }\left(x\right)=x^4\left({\sec }^{\mathrm{2 }}x\right)+\left(4x^3\right)\left(\mathrm{tan }x\right)=x^3\left(x{\mathrm{ sec}}^{\mathrm{2 }}x+4\mathrm{ tan }x\right)$

then by Equation (4.1), above,

${\displaystyle f^{\prime }\left(x\right)=-\frac{\sqrt{3}{x}^3\left(x{\mathrm{ sec}}^{\mathrm{2 }}x+4\mathrm{ tan }x\right)}{{\left(x^4\mathrm{ tan }x\right)}^2}=-\frac{\sqrt{3}\left(x{\mathrm{ sec}}^{\mathrm{2 }}x+4\mathrm{ tan }x\right)}{x^5{ tan}^{\mathrm{2 }}x}\text{.}}$

Example 4.49. For the function ${\displaystyle h\left(x\right)=\frac{1}{\sqrt{{\cot }^{\mathrm{3 }}x-6x}}\text{,}}$ we have, by the general power rule

${\displaystyle \frac{d}{dx}{\cot }^{\mathrm{3 }}x-6x=-3\left({\cot }^{\mathrm{2 }}x{\mathrm{ csc}}^{\mathrm{2 }}x+2\right)\text{.}}$

Hence, by Equation (4.3), above,

${\displaystyle h^{\prime }\left(x\right)=\frac{3\left({\cot }^{\mathrm{2 }}x{\mathrm{ csc}}^{\mathrm{2 }}x+2\right)}{2{\left({\cot }^{\mathrm{3 }}x-6x\right)}^{3∕2}}\text{.}}$

The Leibnitz notation of the derivative plays an important role in the chain rule. If we set $u=g\left(x\right)$, then $f\circ g\left(x\right)=f\left(g\left(x\right)\right)=f\left(u\right)$. In Leibnitz notation

${\displaystyle \frac{df}{du}=f^{\prime }\left(u\right)=f^{\prime }\left(g\left(x\right)\right)\text{ and }\frac{du}{dx}=g^{\prime }\left(x\right)\text{.}}$

Therefore, the chain rule in Leibnitz notation is

${\displaystyle \frac{df}{dx}=\frac{df}{du}\frac{du}{dx}\text{.}}$

Observe in the next examples how effective this notation is.

Example 4.50. For the function in Example 4.49, the function $h\left(x\right)$ is the composition of $f\left(u\right)=u^{-1∕2}$ and $u=g\left(x\right)={cot}^{\mathrm{3 }}x-6x$; that is $h\left(x\right)=f\left(u\right)=f\circ g\left(x\right)={\left({\cot }^{\mathrm{3 }}x-6x\right)}^{-1∕2}$.

In Leibnitz notation,

${\displaystyle \begin{array}{ll}\hfill h^{\prime }\left(x\right)& ={\displaystyle \frac{df}{du}}{\displaystyle \frac{du}{dx}}={\displaystyle \frac{-1}{2}}{u}^{-3/2}\left(-3{\cot }^{2}x{\csc }^{2}x-6\right)\hfill \\ \hfill & ={\displaystyle \frac{3{\cot }^{2}x{\csc }^{2}x+6}{2u^{3/2}}}.\hfill \end{array}}$

Substituting $u\text{,}$ we have ${\displaystyle h^{\prime }\left(x\right)=\frac{3{ cot}^{\mathrm{2 }}x{\mathrm{ csc}}^{\mathrm{2 }}x+6}{2{\left({\cot }^{\mathrm{3 }}x-6x\right)}^{3∕2}}\text{.}}$

Example 4.51. For the function $R\left(x\right)={\left(\sec \left(\sqrt{x}\right)\right)}^{-2}$ in Example 4.47, the outside and inside functions ($f_1\left(u\right)=u^{-2}$ and $u=g_1\left(x\right)=\sec \left(\sqrt{x}\right)\text{,}$ respectively) give the function $R\left(x\right)=f_1\left(u\right)=f_1\circ g_1\left(x\right)\text{.}$

At the same time, $g_1\left(x\right)$ is the composition of $f_2\left(t\right)=\mathrm{sec }t$ and $t=g_2\left(x\right)=\sqrt{x}\text{;}$ that is, $g_1\left(x\right)=f_2\left(t\right)=f_2\circ g_2\left(x\right)\text{.}$ Hence, $R\left(x\right)=f_1\circ \left(f_2\circ g_2\left(x\right)\right)\text{.}$

The derivative in Leibnitz notation is

${\displaystyle R^{\prime }\left(x\right)=\frac{d{f}_1}{du}\frac{du}{dt}\frac{dt}{dx}=-2{\left(u\right)}^{-3}\left(\mathrm{sec }t\mathrm{ tan }t\right)\frac{1}{2\sqrt{x}}=\frac{-\mathrm{sec }t\mathrm{ tan }t}{u^3\sqrt{x}}\text{.}}$

Substituting $u$ and $t\text{,}$ we conclude that

${\displaystyle R^{\prime }\left(x\right)=\frac{-\sec \left(\sqrt{x}\right)\tan \left(\sqrt{x}\right)}{{\sec }^3\left(\sqrt{x}\right)\sqrt{x}}=-\frac{\tan \left(\sqrt{x}\right)}{{\sec }^2\left(\sqrt{x}\right)\sqrt{x}}\text{.}}$

Tutorial 1: Applying Differentiation Rules

Tutorial 2: Differentiation

Tutorial 3: Differentiation