Calculus Solutions Ex#2.5

 

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Fomulas: Derivatives of trigonometric functions
$$ \begin{array}{ll} \frac{d}{d x}[\sin x]=\cos x & \frac{d}{d x}[\cos x]=-\sin x \\ \\ \frac{d}{d x}[\tan x]=\sec ^2 x & \frac{d}{d x}[\sec x]=\sec x \tan x \\ \\ \frac{d}{d x}[\cot x]=-\csc x^2 & \frac{d}{d x}[\csc x]=-\csc x \cot x \end{array} $$
Q.01 Find $f^{\prime}(x)$ where $f(x)=4 \cos x+2 \sin x$
Solution:
$$f(x)=4 \cos x+2 \sin x$$ differentiating w.r.t $x$ $$ \begin{aligned} & f^{\prime}(x)=\frac{d}{d x}[4 \cos x+2 \sin x]=4 \frac{d}{d x}[\cos x]+2 \frac{d}{d x}[\sin x] \\ \\ & f^{\prime}(x)=-4 \sin x+2 \cos x \end{aligned} $$
Q.02 Find $f^{\prime}(x)$ where $$f(x)=5 / x^2+\sin x=5 x^{-2}+\sin x$$
Solution:
$$f(x)=5 / x^2+\sin x=5 x^{-2}+\sin x$ differentiating w.r.t $x$ \\ \\ & \begin{aligned} & f^{\prime}(x)=\frac{d}{d x}\left[5 x^{-2}+\sin x\right]. \\ \\ & f^{\prime}(x)=5 \frac{d}{d x}\left[x^{-2}\right]+\frac{d}{d x}[\sin x] \\ \\ & =5(-2) x^{-2-1}+\cos x \\ \\ & =-10 x^{-3}+\cos x \end{aligned} $$
Q.03 Find $f^{\prime}(x)$ where $f(x)=-4 x^2 \cos x$
Solution:
$$f(x)=-4 x^2 \cos x $$ differentiating w.r.t $x$ $$\\ \\ & \begin{aligned} f^{\prime}(x) & =-4 \frac{d}{d x}\left[x^2 \cos x\right]=-4\left[\frac{d}{d x}\left(x^2 \cos x\right)\right] \\ \\ & =-4\left[x^2 \frac{d}{d x}(\cos x)+\cos x \frac{d}{d x}\left[x^2\right)\right] \quad \text { (product rule) } \end{aligned} $$ $$ \begin{aligned} & f^{\prime}(x)=-4\left[x^2(-\sin x)+(\cos x)\left(2 x^{2-1} \right)\right] \\ \\& f^{\prime}(x)=-4\left[-x^2 \sin x+2 x \cos x\right]=4 x^2 \sin x-8 x \cos x \end{aligned} $$
Q.04 Find $f^{\prime}(x)$ where $f(x)=2 \sin ^2 x$
Solution:
$$f(x)=2 \sin ^2 x$$ differentiating wr.t $x$ (by using Chain rule) $$  \begin{aligned} & f^{\prime}(x)=2 \frac{d}{d x}\left[\sin ^2 x\right]=2\left[2 \sin x \frac{d}{d x}[\sin x]\right]. \\ \\ & =4 \sin x \cos x \\ \\& \end{aligned} $$ Chain rule: $$ \begin{aligned} & \frac{d}{d x}[g(x)]^n =n g(x)^{n-1} \frac{d g}{d x} \end{aligned} $$
Q.05 Find $f^{\prime}(x)$ where $ f(x)=\frac{5-\cos x}{5+\sin x}$
Solution:
$$ f(x)=\frac{5-\cos x}{5+\sin x}$$ differentiating w.r.t $x$ (using quotient rule) $$ \begin{aligned} & f^{\prime}(x)=\frac{d}{d x}\left[\frac{5-\cos x}{5+\sin x}\right] \\ \\ & f^{\prime}(x)=\frac{[5+\sin x] \frac{d}{d x}[5-\cos x]-[5-\cos x] \frac{d}{d x}[5+\sin x]}{(5+\sin x)^2} \\ \\ & f^{\prime}(x)=\frac{[5+\sin x]\left[\frac{d(5)}{d x}-\frac{d}{d x}(\cos x)\right]-[5-\cos x]\left[\frac{d}{d x}(5)+\frac{d}{d x}(\sin x)\right]}{(5+\sin x)^2} \\ \\& f^{\prime}(x)=\frac{[5+\sin x][0-(-\sin x)]-[5-\cos x][0+\cos x]}{(5+\sin x)^2} \\ \\ & \because \frac{d}{d x}\left[sin x\right]=cos x  \\ \\ & \frac{d}{d x}[\cos x]=-\sin x \\ \\  & f^{\prime}(x)=\frac{(5+\sin x)(\sin x)-(5-\cos x)(\cos x)}{(5+\sin x)^2} \\ \\ & \end{aligned} $$ $$ \begin{aligned} & f^{\prime}(x)=\frac{5 \sin x+\sin ^2 x-5 \cos x+\cos ^2 x}{(5+\sin x)^2} \\ \\& f^{\prime}(x)=\frac{5 \sin x-5 \cos x+\sin ^2 x+\cos ^2 x}{(5+\sin x)^2} \\ \\ & f^{\prime}(x)=\frac{5 \sin x-5 \cos x+1}{(5+\sin x)^2} \quad \therefore \sin ^2 x+\cos ^2 x=1 \end{aligned} $$
Q.06 Find $f^{\prime}(x)$ where $ f(x)=\frac{\sin x}{x^2+\sin x} $
Solution:
$$ f(x)=\frac{\sin x}{x^2+\sin x} $$ differentiating w.r.t. $x$ (using quotient rule) $$ \begin{aligned} & f^{\prime}(x)=\frac{d}{d x}\left[\frac{\sin x}{\left.x^2+\sin x\right}\right]. \\ \\ & f^{\prime}(x)=\frac{\left[x^2+\sin x\right] \frac{d}{d x}[\sin x]-\sin x \frac{d}{d x}\left[x^2+\sin x\right]}{\left(x^2+\sin x\right)^2} \\ \\& f^{\prime}(x)=\frac{\left.\left[x^2+\sin x\right][\cos x]-\sin x\left[\frac{d}{d x}\left(x^2\right)+\frac{d}{d x} \sin x\right)\right]}{\left(x^2+\sin x\right)^2} \\ \\& \therefore \frac{d}{d x} (\sin x )=\cos x \\ \\& f^{\prime}(x)=\frac{x^2 \cos x+\cos x \sin x-\sin x[2 x+\cos x]}{\left(x^2+\sin x\right)^2} \\ \\ & f^{\prime}(x)=\frac{x^2 \cos x+\cos x \sin x-2 x \sin x-\cos x \sin x}{\left(x^2+\sin x\right)^2} \\ \\& f^{\prime}(x)=\frac{x^2 \cos x-2 x \sin x}{\left(x^2+\sin x\right)^2} \end{aligned} $$