Chapter 13: Limits And Derivatives Miscellaneous Exercise

Ans: (i) Let $f(x)=-x$. Accordingly, $f(x+h)=-(x+h)$

By first principle, $f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{-(x+h)-(-x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{-x-h+x}{h}$

$=\lim _{n \rightarrow 0} \dfrac{-h}{h}$

$=\lim _{h \rightarrow 0}(-1)=-1$

(ii) Let $f(x)=(-x)^{-1}=\dfrac{1}{-x}=\dfrac{-1}{x} .$ Accordingly, $f(x+h)=\dfrac{-1}{(x+h)}$

By first principle,

$f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-1}{(x+h)}-\left(\dfrac{-1}{x}\right)\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-x+(x+h)}{x(x+h)}\right]$

$=\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{h}{x(x+h)}\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{x(x+h)}$

$=\dfrac{1}{x \cdot X}=\dfrac{1}{x^{2}}$

(iii) Let $f(x)=\sin (x+1)$. Accordingly, $f(x+h)=\sin (x+h+1)$

By first principle,

$f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}[\sin (x+h+1)-\sin (x+1)]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{x+h+1+x+1}{2}\right) \sin \left(\dfrac{x+h+1-x-1}{2}\right)\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{2 x+h+2}{2}\right) \sin \left(\dfrac{h}{2}\right)\right]$

$=\lim _{n \rightarrow 0}\left[\cos \left(\dfrac{2 x+h+2}{2}\right) \cdot \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right]$

$\begin{array}{l}=\lim _{n \rightarrow 0} \dfrac{1}{h} \cos \left(\dfrac{2 x+h+2}{2}\right) \cdot \lim _{\dfrac{b}{2} \rightarrow} \dfrac{\sin \left(\dfrac{h}{2}\right)}{h} \dfrac{h}{\left.\dfrac{h}{2}\right)} \quad\left[\text { As } h \rightarrow 0 \Rightarrow \dfrac{h}{2} \rightarrow 0\right]\\=\cos \left(\dfrac{2 x+0+2}{2}\right) \cdot 1 \quad\left[\lim _{h \rightarrow 0} \dfrac{\sin x}{x}=1\right]\\=\cos (x+1)\\\text (iv) Let f(x)=\cos \left(x-\dfrac{\pi}{8}\right) Accordingly, f(x+h)-\cos \left(x+h-\dfrac{\pi}{8}\right)\\\text { By first principle, }\\f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+f)-f(x)}{h}\\=\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\cos \left(x+h-\dfrac{\pi}{8}\right)-\cos \left(x-\dfrac{\pi}{8}\right)\right]\\=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[-2 \sin \dfrac{\left(x+h-\dfrac{\pi}{8}+x-\dfrac{\pi}{8}\right)}{2} \sin \left(\dfrac{x+h-\dfrac{\pi}{8}-x+\dfrac{\pi}{8}}{2}\right)\right]\\=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[-2 \sin \left(\dfrac{2 x+h-\dfrac{\pi}{4}}{2}\right) \sin \left(\dfrac{h}{2}\right)\right]\\=\lim _{n \rightarrow 0}\left[-\sin \left(\dfrac{2 x+h-\dfrac{\pi}{4}}{2}\right) \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right]\\\left.=\lim _{n \rightarrow 0}\left[-\sin \left(\dfrac{2 x+h-\dfrac{\pi}{4}}{2}\right)\right] \cdot \lim _{\dfrac{\pi}{2} \rightarrow 0} \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)} \quad \text { [As } h \rightarrow 0 \Rightarrow \dfrac{h}{2} \rightarrow 0\right]\\=-\sin \left(\dfrac{2 x+0-\dfrac{\pi}{4}}{2}\right) \cdot 1\end{array}$

$=-\sin \left(x-\dfrac{\pi}{8}\right)$

Ans: Let $f(x)=x+a$. Accordingly. $f(x+h)-x+h+a$ By first principle,

$f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$\lim _{n \rightarrow 0} \dfrac{x+h+a-x-a}{h}$

$\lim _{n \rightarrow 0}\left(\dfrac{h}{h}\right)$

$-\lim _{n \rightarrow 0}(1)$

$=1$

Ans: Let $f(x)=(p x+q)\left(\dfrac{r}{x}+s\right)$

By Leibnitz product rule.

$f^{\prime}(x)=(p x+q)\left(\dfrac{r}{x}+s\right)^{\prime}+\left(\dfrac{r}{x}+s\right)(p x+q)^{\prime}$

$-(p x+q)\left(r x^{-1}+s\right)^{\prime}+\left(\dfrac{r}{x}+s\right)(p)$

$-(p x+q)\left(-n x^{2}\right)+\left(\dfrac{r}{x}+s\right) p$

$=(p x+q)\left(\dfrac{-r}{x^{2}}\right)+\left(\dfrac{r}{x}+s\right) p$

$=\dfrac{-p x}{x}-\dfrac{q r}{x^{2}}+\dfrac{p r}{x}+p s$

$p s-\dfrac{q r}{x^{2}}$

Ans: Let $f^{\prime}(x)=(a x+b)(c x+d)^{2}$

By Leibnitz product rule,

$f^{\prime}(x)=(a x+b) \dfrac{d}{d x}(c x+d)^{2} \dfrac{d}{d x}(a x+b)$

$(a x+b) \dfrac{d}{d x}\left(c^{2} x^{2}+2 c d x^{2}\right)+(c x+d)^{2} \dfrac{d}{d x}(a x+b)$

$(a x+b)\left[\dfrac{d}{d x}\left(c^{2} x^{2}\right)+\dfrac{d}{d x}(2 c d x)+\dfrac{d}{d x} d^{2}\right]+(c x+d)^{2}\left[\dfrac{d}{d x} a x+\dfrac{d}{d x} b\right]$

$=(a x+b)\left(2 c^{2} x+2 c d\right)+(c x+d)^{2} a$

$-2 c(a x+b)(c x+d)+a(c x+d)^{2}$

Ans: Let $f(x)=\dfrac{a x+b}{c x+d}$

By quotient rule,

$f(x)=\dfrac{(c x+d) \dfrac{d}{d x}(a x+b)-(a x+b) \dfrac{d}{d x}(c x+d)}{(c x+d)^{2}}$

$=\dfrac{(c x+d)(a)-(a x+d)(c)}{(c x+d)^{2}}$

$\dfrac{a c x+a d-a c x-b c}{(c x+d)^{2}}$

$\dfrac{a d-b c}{(c x+d)^{2}}$

Ans: Let$f(x)=\dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x}}=\dfrac{\dfrac{x+1}{x}}{\dfrac{x-1}{x}}=\dfrac{x+1}{x-1}$, where $x \neq 0$

By quotient rule, $f^{\prime}(x)=\dfrac{(x-1) \dfrac{d}{d x}(x-1)-(x+1) \dfrac{d}{d x}(x-1)}{(x-1)^{2}}, x \neq 0,1$

$=\dfrac{(x-1)(1)-(x+1)(1)}{(x-1)^{2}}, x \neq 0,1$

$\dfrac{x-1-x-1}{(x-1)^{2}}, x \neq 0,1$

$\dfrac{-2}{(x-1)^{2}}, x \neq 0,1$

Ans: Let $f(x)=\dfrac{1}{a x^{2}+b x+c}$

By quotient rule,

$f^{\prime}(x)=\dfrac{\left(a x^{2}+b x+c\right) \dfrac{d}{d x}(1)-\dfrac{d}{d x}\left(a x^{2}+b x+c\right)}{\left(a x^{2}+b x+c\right)^{2}}$

$\dfrac{\left(a x^{2}+b x+c\right)(0)-(2 a x+b)}{\left(a x^{2}+b x+c\right)^{2}}$

$\dfrac{-(2 a x+b)}{\left(a x^{2}+b x+c\right)^{2}}$

Ans: Let $f(x)=\dfrac{a x+b}{p x^{2}+q x+r}$

By quotient rule,

$f^{\prime}(x)=\dfrac{\left(p x^{2}+q x+r\right) \dfrac{d}{d x}(a x+b)-(a x+b) \dfrac{d}{d x}\left(p x^{2}+q x+r\right)}{\left(p x^{2}+q x+r\right)^{2}}$

$=\dfrac{\left(p x^{2}+q x+r\right)(a)-(a x+b)(2 p x+q)}{\left(p x^{2}+q x+r\right)^{2}}$

$=\dfrac{a p x^{2}+a q x+a r-a q x+2 n p x+b q}{\left(p x^{2}+q x+r\right)^{2}}$

$\dfrac{-a p x^{2}+2 b p x+a r-b q}{\left(p x^{2}+q x+r\right)^{2}}$

Ans: Let $f(x)=\dfrac{p x^{2}+q x+r}{a x+b}$

By the quotient rule,

$\dot{f}(x)=\dfrac{(a x+b) \dfrac{d}{d x}\left(p x^{2}+q x+\eta\right)-\left(p x^{2}+q x+r\right) \dfrac{d}{d x}(a x+b)}{(a x+b)^{2}}$

$\dfrac{(a x+b)(2 p x+q)-\left(p x^{2}+q x+r\right)(a)}{(a x+b)^{2}}$

$=\dfrac{2 a p x^{2}+a q x+2 b p x+b q-a q x^{2}-a q x-a r}{(a x+b)^{2}}$

$\dfrac{a p x^{2}+2 b p x+b q-a r}{(a x+b)^{2}}$

Ans: Let $f(x)=\dfrac{a}{x^{4}}-\dfrac{b}{x^{2}}+\cos x$

$f^{\prime}(x)=\dfrac{d}{d x}\left(\dfrac{a}{x^{4}}\right)-\dfrac{d}{d x}\left(\dfrac{a}{x^{2}}\right)+\dfrac{d}{d x}(\cos x)$

$a \dfrac{d}{d x}\left(x^{-4}\right)-b \dfrac{d}{d x}\left(x^{2}\right)+\dfrac{d}{d x}(\cos x)$

$-a\left(-4 x^{-5}\right)-b\left(-2 x^{3}\right)+(-\sin x) \quad\left[\dfrac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right.$ and $\left.\dfrac{d}{d x}(\cos x)=-\sin x\right]$

$\dfrac{-4 a}{x^{5}}+\dfrac{2 b}{x^{3}}-\sin x$