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

$$ \begin{aligned} & \text { Since } \quad f(x)=\ln (2 x+1) \\ \\ & \text { Differentiating w.r.t } x \\ \\ & f^{\prime}(x)=\frac{d}{d x}[\ln (2 x+1)] \\ \\ & f^{\prime}(x)=\frac{1}{2 x+1} \frac{d}{d x}[2 x+1] \\ \\ & f^{\prime}(x)=\frac{1}{2 x+1}(2(1)+0) \\ \\ & f^{\prime}(x)=\frac{2}{2 x+1} \\ \\ & f^{\prime}\left(f^{-1}(x)\right)=\frac{2}{2 f^{-1} \cdot(x)+1} \\ \\ & f^{\prime}\left(f^{-1}(x)\right)=\frac{2}{2\left[\frac{1}{2}\left(e^{x}-1\right)\right]+1} \\ \\ & =\frac{2}{e^{x}-1+1}=2 / e^{x} \\ \\ & \therefore\left(f^{-1}\right)^{\prime}(x)=\frac{1}{f^{\prime}\left(f^{-1}(x)\right)}=1 / 2 / e^{x} \\ \\ & =e^{x} / 2 \end{aligned} $$
Q.05 Determine which the function $f$ in one to one by examining the sign of $f^{\prime}(x)$ .
Solution:
Note: If $f^{\prime}(x)>0$ or $f^{\prime}(x)<0$ on some open interval then $f$ is one to one.

$$ \begin{enumerate} \setcounter{enumi}{4} \item (a) $\quad \quad f(x)=x^{2}+8 x+1$ \\ \\ Differentiating w.t.t $x$ \end{enumerate} $$

$$ f^{\prime}(x)=2 x+8 $$

$$ set \quad f^{\prime}(x)=0$$

$$ \Rightarrow \quad 2 x+8 & =0 $$ $$ x & =-4 $$

$$ \begin{aligned} & f^{\prime}(x)<0$ on $(-\infty,-4) \\ \\ & \text {and} \quad f^{\prime}(x)>0$ on $(-4, \infty)\end{aligned}$$ we can not determine by this information whether $f$ is one to one or not. WE also see that $$f(1)=(1)^2+8(1)+1=1+8+1=10$$ $$f(-9)=(-9)^2+ 8(-9)+1=81-72+1=10$$ $$ \Rightarrow f(1)=10=f(-9) \text { but } 1 \neq-9 $$ thus we have reached to the conclusion that $f$ is not one to one.

(b) $$ \quad f(x)=2 x^{5}+x^{3}+3 x+2$$

Differentiating w.r.t $x$ $$f^{\prime}(x)=\frac{d}{d x}\left[2 x^{5}+x^{3}+3 x+2\right]$$

$$f^{\prime}(x)=2 \frac{d}{d x}\left[x^{5}\right]+\frac{d}{d x}\left[x^{3}\right]+3 \frac{d}{d x}[x]+\frac{d}{d x}[2]$$

$$f^{\prime}(x)=2\left(5 x^{5-1}\right)+3 x^{3-1}+3(1)+0$$

$$f^{\prime}(x)=10 x^{4}+3 x^{2}+3$$ which is greater than zero for all $x$ $\Rightarrow f^{\prime}(x)>0$, for all $x$ $\Rightarrow f$ is one to one.