Calculus Solutions Ex# 3.2

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Formulas: Derivatives of logarithmic functions.
 
$$ \begin{aligned} & \text { (1) } \frac{d}{d x}[\ln x]=\frac{1}{x}, \quad x>0 \\ \\ & \text { (2) } \frac{d}{d x}\left[\log _{b}^{x}\right]=\frac{1}{x \ln b}, x>0 \end{aligned} $$ Generalized derivative formulas. $$ \begin{aligned} & \text { (1) } \frac{d}{d x}[\ln u]=\frac{1}{u} \frac{d u}{d x} \\ \\ & \text { (2) } \frac{d}{d x}\left[\log _{b} u\right]=\frac{1}{u \ln b} \frac{d u}{d x} \end{aligned} $$ $$ \text { (3) }\frac{d}{d x}[\ln |x|]=\frac{1}{x} , \text { if } x \neq 0 $$
Q.01 Find $\frac{d y}{d x}$ where $y=\ln 5 x$.
Solution:
$$ \quad y=\ln 5 x $$ Differentiating w.r.t. $x$ $$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}[\ln 5 x] \\ \\ \frac{d y}{d x} & =\frac{1}{5 x} \frac{d(5 x)}{d x} \\ \\ \frac{d y}{d x} & =\frac{1}{5 x}(5)(1)=\frac{5}{5 x} \\ \\ \frac{d y}{d x} & =\frac{1}{x} \end{aligned} $$
Q.02 Find $\frac{d y}{d x}$ where $ y=\ln (x / 3)$.
Solution:
$$ \quad y=\ln (x / 3)$$ Differentiating w.r.t $x$ $$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}[\ln (x / 3)] \\ \\ & \frac{d y}{d x}=\frac{1}{x / 3} \cdot \frac{d}{d x}(x / 3)=\frac{1}{(x / 3)} \cdot \frac{1}{3}(1) \\ \\ & \frac{d y}{d x}=\frac{1}{3 \cdot x / 3}=\frac{1}{x} \end{aligned} $$
Q.03 Find $\frac{d y}{d x}$ where $ y=\ln |1+x|$.
Solution:
$$ y=\ln |1+x|$$ $$ \begin{aligned} & \text { Differentiating w.r.t } x \\ \\ & \frac{d y}{d x}=\frac{d}{d x}[\ln |1+x|] \\ \\ & \frac{d y}{d x}=\frac{1}{1+x} \frac{d(1+x)}{d x} \\ \\ & \text {because} \frac{d}{d x} \ln |x|=\frac{1}{x} \text { ,if } x \neq 0 \\ \\ & \therefore \frac{d y}{d x}=\frac{1}{1+x}(0+1)=\frac{1}{d x} \end{aligned} $$
Q.04 Find $\frac{d y}{d x}$ where $ y=\ln (2+\sqrt{x}$.
Solution:
$$ y=\ln (2+\sqrt{x}$$ Differentiating w.r.t $x$ $$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}[\ln (2+\sqrt{x})] \\ \\ & \frac{d y}{d x}=\frac{1}{2+\sqrt{x}} \frac{d}{d x}[2+\sqrt{x}] \\ \\ & \frac{d y}{d x}=\frac{1}{2+\sqrt{x}}\left[0+\frac{d}{d x} x^{1 / 2}\right] \\ \\ & \frac{d y}{d x}=\frac{1}{2+\sqrt{x}}\left[\frac{1}{2} x^{\frac{1}{2}-1}\right] \\ \\ & \frac{d y}{d x}=\frac{1}{2+\sqrt{x}} \cdot \frac{1}{2} x^{-1 / 2} \\ \\ & \frac{d y}{d x}=\frac{1}{2 x^{1 / 2}(2+\sqrt{x})} \\ \\ & \frac{d y}{d x}=\frac{1}{2 \sqrt{x}(2+\sqrt{x})}=\frac{1}{4 \sqrt{x}+2(\sqrt{x})^{2}}=\frac{1}{4 \sqrt{x}+2 x} \end{aligned}$$
Q.05 Find $\frac{d y}{d x}$ where $ y=\ln \left|x^{2}-1\right|$.
Solution:
$$ y=\ln \left|x^{2}-1\right|$$ differentiating w.r.t $x$ $$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left[\ln \left|x^{2}-1\right|\right] \\ \\ & \frac{d y}{d x}=\frac{1}{x^{2}-1} \frac{d}{d x}\left(x^{2}-1\right) \\ \\ & \frac{d y}{d x}=\frac{1}{x^{2}-1}\left(2 x^{2-1}-0\right)=\frac{2 x}{x^{2}-1} \end{aligned} $$
Q.06 Find $\frac{d y}{d x}$ where $ y=\ln \left|x^{3}-7 x^{2}-3\right|$.
Solution:
$$y=\ln \left|x^{3}-7 x^{2}-3\right|$$ differentiating w.r.t $x$ $$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left[\ln \left|x^{3}-7 x^{2}-3\right|]. \\ \\ & \frac{d y}{d x}=\frac{1}{x^{3}-7 x^{2}-3} \cdot \frac{d}{d x}\left[x^{3}-7 x^{2}-3\right] \\ \\ & \frac{d y}{d x}=\frac{1}{x^{3}-7 x^{2}-3} \cdot\left[3 x^{3-1}-7\left(2 x^{2-1}\right)-0\right] \\ \\ & \frac{d y}{d x}=\frac{1}{x^{3}-7 x^{2}-3} \cdot\left[3 x^{2}-14 x\right]=\frac{3 x^{2}-14 x}{x^{3}-7 x^{2}-3} \end{aligned} $$
Q.07 Find $\frac{d y}{d x}$ where $ y=\ln \left(\frac{x}{x^{2}+1}\right)$.
Solution:
$$y=\ln \left(\frac{x}{x^{2}+1}\right)$$ differentiating w.r.t $x$ $$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left[\ln \left(\frac{x}{1+x^{2}}\right)\right] \quad \because \ln \left(\frac{x}{y}\right)=\ln x-\ln y \\ \\ & \frac{d y}{d x}=\frac{d}{d x}\left[\ln x-\ln \left(1+x^{2}\right)\right] \\ \\ & \frac{d y}{d x}=\frac{d}{d x}[\ln x]-\frac{d}{d x}\left[\ln \left(1+x^{2}\right)\right] \\ \\ & \frac{d y}{d x}=\frac{1}{x}-\frac{1}{1+x^{2}} \frac{d}{d x}\left(1+x^{2}\right) \\ \\ & \frac{d y}{d x}=\frac{1}{x}-\frac{1}{1+x^{2}}\left(0+2 x^{2-1}\right) \\ \\ & \frac{d y}{d x}=\frac{1}{x}-\frac{2 x}{1+x^{2}} \\ \\ & \frac{d y}{d x}=\frac{1+x^{2}-2 x^{2}}{x\left(1+x^{2}\right)} =\frac{1-x^{2}}{x\left(1+x^{2}\right)} \end{aligned} $$