Calculus Solutions Ex# 3.3

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Formulas: Derivatives of exponential and inverse trigonometric functions functions.

$\begin{aligned} & \text { (1) }\left(f^{-1}\right)^{\prime}(x)=\frac{1}{f^{\prime}\left(f^{-1}(x)\right)} \text { or for } x=f(y) , \frac{d y}{d x}=\frac {1}{dx/dy} \\ \\ & \text { (2) } \frac{d}{d x}\left[b^{x}\right]=b^{x} \ln b \\ \\ & \text { (3) } \frac{d}{d x}\left[e^{x}\right]=e^{x} \end{aligned}$

If $u$ is a differentiable function of $x$

$\begin{aligned} & \text { (4) } \quad \frac{d}{d x}\left[b^{u}\right]=b^{u} \ln b \frac{d u}{d x} \\ \\ & \text { (5) } \quad \frac{d}{d x}[e^{u}]=e^{u} \frac{d u}{d x} \\ \\ & \text { (6) } \quad \frac{d}{d x}\left[\sin^{-1}u \right]=\frac{1}{\sqrt{1-u^{2}}} \frac{d u}{d x} \\ \\ & \text { (7) } \quad \frac{d}{d x}\left[\cos ^{-1} u\right]=-\frac{1}{\sqrt{1-u^{2}}} \frac{d u}{d x} \\ \\ & \text { (8) } \quad \frac{d}{d x}\left[\tan ^{-1} u\right]=\frac{1}{1+u^{2}} \frac{d u}{d x} \\ \\ & \text { (9) } \quad \frac{d}{d x}\left[\cot ^{-1} u\right]=-\frac{1}{1+u^{2}} \frac{d u}{d x} \\ \\ & \text { (10) } \quad \frac{d}{d x}\left[\sec ^{-1} u\right]=\frac{1}{|u| \sqrt{u^{2}-1}} \frac{d u}{d x} \\ \\ & \text { (11) } \quad \frac{d}{d x}\left[\csc ^{-1} u\right]=-\frac{1}{|u| \sqrt{u^{2}-1}} \frac{d u}{d x} \end{aligned}$

Q.03 Find $\left(f^{-1}\right)^{\prime}(x)$ .

Solution:

$\quad f(x)=\frac{2}{x+3}$

$\begin{aligned} & \text {Let} \quad y=\frac{2}{x+3} \\ \\ & (x+3) y=2 \\ \\ & x+3=2 / y \\ \\ & x=\frac{2}{y}-3 \\ \\ & x=f^{-1}(y)=\frac{2}{y}-3 \\ & \text {Interchanging} \quad x \quad \text {and} \quad y \\ \\ & f^{-1}(x)=\frac{2}{x}-3 \end{aligned}$

Differentiating w.r.t $x$

$\begin{aligned} & \left(f^{-1}\right)^{\prime}(x)=\frac{d}{d x}\left[\frac{2}{x}-3\right]=\frac{d}{d x}[2 / x]-\frac{d}{d x}[3] \\ \\ & \left(f^{-1}\right)^{\prime}(x)=\frac{d}{d x}[2 x^{-1}] \\ \\ & \left(f^{-1}\right)^{\prime}(x)= 2 (-1) x^{-1-1}=-2 x^{-2} \end{aligned}$

Now using formula $\quad \left(f^{-1}\right)^{\prime}(x)=\frac{1}{f^{\prime}\left(f^{-1}(x)\right)}$

$\begin{aligned} & f(x)=\frac{2}{x+3}=2(x+3)^{-1} \\ \\ & \text { Differentiating w.r.t } x \\ \\ & f^{\prime}(x)=2\left(-1(x+3)^{-1-1}\right)=-2(x+3)^{-2}=-\frac{2}{(x+3)^{2}} \\ \\ & f^{\prime}(x)=-\frac{2}{(x+3)^{2}} \end{aligned}$

$\begin{aligned} & \text { Formula: } \quad\left(f^{-1}\right)^{\prime}(x)=\frac{1}{f^{\prime}\left(f^{-1}(x)\right)} \\ \\ & \left(f^{-1}\right)^{\prime}(x)=\frac{1}{-2 /\left(f^{-1}(x)+3\right)^{2}}=-\frac{\left(f^{-1}(x)+3\right)^{2}}{2} \\ \\ & \left(f^{-1}\right)^{\prime}(x)=\frac{-\left(\frac{2}{x}-3+3\right)^{2}}{2} \quad \therefore f^{-1}(x)=\frac{2}{x}-3 \\ \\ & \left(f^{-1}\right)^{\prime}(x)=-\frac{(2 / x)^{2}}{2}=-\frac{4}{2 x^{2}} \\ \\ & \left(f^{-1}\right)^{\prime}(x)=-2 / x^{2}$

Q.04 Find $\left(f^{-1}\right)^{\prime}(x)$ .

Solution:

$\begin{aligned} & \quad f(x)=\ln (2 x+1) \\ \\ & y=\ln (2 x+1) \\ \\ & e^{y}=2 x+1 \\ \\ & 2 x=e^{y}-1 \\ \\ & x=\frac{1}{2}\left(e^{y}-1\right) \\ \\ & x=f^{-1}(y)=\frac{1}{2}\left(e^{y}-1\right) \\ \\ & \text { interchanging } x \text { and } y \\ \\ & f^{-1}(x)=\frac{1}{2}\left(e^{x}-1\right) \\ \\ & \text { Differentiating w. r. t } x \\ \\ & \left(f^{-1}\right)^{\prime}(x)=\frac{1}{2} \frac{d}{d x}\left[e^{x}-1\right]=\frac{1}{2}\left(e^{x}-0\right)=e^{x} / 2 \end{aligned}$

Now using formula $\left(f^{-1}\right)^{\prime}(x)=\frac{1}{f^{\prime}\left(f^{-1}(x)\right)}$

$\begin{aligned} & \text { Since } \quad f(x)=\ln (2 x+1) \\ \\ & \text { Differentiating w.r.t } x \\ \\ & f^{\prime}(x)=\frac{d}{d x}[\ln (2 x+1)] \\ \\ & f^{\prime}(x)=\frac{1}{2 x+1} \frac{d}{d x}[2 x+1] \\ \\ & f^{\prime}(x)=\frac{1}{2 x+1}(2(1)+0) \\ \\ & f^{\prime}(x)=\frac{2}{2 x+1} \\ \\ & f^{\prime}\left(f^{-1}(x)\right)=\frac{2}{2 f^{-1} \cdot(x)+1} \\ \\ & f^{\prime}\left(f^{-1}(x)\right)=\frac{2}{2\left[\frac{1}{2}\left(e^{x}-1\right)\right]+1} \\ \\ & =\frac{2}{e^{x}-1+1}=2 / e^{x} \\ \\ & \therefore\left(f^{-1}\right)^{\prime}(x)=\frac{1}{f^{\prime}\left(f^{-1}(x)\right)}=1 / 2 / e^{x} \\ \\ & =e^{x} / 2 \end{aligned}$