Limits and Derivatives

58. Given f (x) = cosec x. cot x.

By Leibnitz product rule,

So, g(x) = $\underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h}$

$=\underset{h\to 0}{lim}\frac{cot(x+h)-cotx}{h}$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{cos(x+h)}{sin(x+h)}-\frac{cosx}{sinx}\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{sinxcos(x+h)-cosxsin(x+h)}{sinxsin(x+h)}\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{sin\left(x-(x+h)\right)}{sinxsin(x+h)}\right]$ $\left[\begin{array}{cc}\hfill ?csin(A-B)=\hfill & \hfill sinAcosB-cosAsinB\hfill \end{array}\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{sin(-h)}{sinx.sin(x+h)}\right]$

$=\underset{h\to 0}{lim}\frac{1}{sinxsin(x+h)}*(-1)\underset{h\to 0}{lim}\frac{sinh}{h}$

$=\frac{1}{sinxsin(x+0)}*(-1)$

= -cosec²x.______(2)

And hx) = $\underset{h\to 0}{lim}\frac{h(x+h)-h(x)}{h}$

$=\underset{h\to 0}{lim}\frac{cosec(x+h)-cosecx}{h}$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{1}{sin(x+h)}-\frac{1}{sinx}\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{sinx-sin(x+h)}{sinx·sin(x+h)}\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}[\frac{2cos\left(\frac{x+x+h}{2}\right)sin\left(\frac{x-(x+h)}{2}\right)}{sinxsin(x+h).}]$

57. Given, f (x) = sin (x + a)

So, f (x) = $\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

$=\underset{h\to 0}{lim}sin\frac{(x+h+a)-sin(x+a)}{h}$

$=\underset{h\to 0}{lim}\frac{1}{h}·2cos\left(\frac{x+h+a+x+a}{2}\right)sin\left(\frac{x+h+a-(x+a)}{2}\right)$

$=\underset{h\to 0}{lim}cos\left(\frac{2x+2a+h}{2}\right)\underset{h\to 0}{lim}\frac{sin(\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.)}{\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$=cos\frac{(2x+2a+0)}{2}*1$

= cos (x + a)

56. Given, f (x) = (ax + b)ⁿ (cx + d)^m

So, f'(x) = ${(ax+b)}^n\frac{d}{dx}{(cx+d)}^n+{(cx+d)}^m\frac{d}{dx}{(ax+b)}^n$

$={(ax+b)}^x·mc{(cx+d)}^{m-1}+{(cx+d)}^{m}na{(ax+b)}^{x-1}$

$={(ax+b)}^{n-1}(cx+d){}^{m-1}[(ax+b)·mc+(cx+d)na]$

$={(ax+b)}^{n-1}(cx+d){}^{m-1}[mc(ax+b)+na(cx+d)]$

55. Given, f (x) = (ax + b)ⁿ

Chain rule,  $\frac{d}{dx}u{(x)}^n=nu{(x)}^{n-1}\frac{du}{dx}u(x)$ where

u (x) is a function of x.

So, f (x) = $\frac{d}{dx}{(ax+b)}^n$

$=n{(ax+b)}^{n-1}\frac{d}{dx}(ax+b)$

$=na{(ax+b)}^{n-1}$

53. Given, f (x) = $=\frac{a}{x^4}-\frac{b}{x}+cosx$

So, f(x) = $\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{a}{{(x+n)}^4}-\frac{b}{{(x+h)}^2}+cos(x+h)-\left(\frac{a}{x^4}-\frac{b}{x^2}+cosx\right)\right]$

$=\underset{h\to 0}{lim}\frac{a}{h}\left[\frac{1}{{(x+h)}^4}-\frac{1}{x^4}\right]-\underset{h\to 0}{lim}\frac{b}{h}\left[\frac{1}{{(x+h)}^2}-\frac{1}{x^2}\right]+\underset{h\to 0}{lim}\frac{1}{h}[cos(x+h)-cosx]$

$=\underset{h\to 0}{lim}\frac{a}{h}\left[\frac{x^4-{(x+h)}^4}{{(x+h)}^4·x^4}\right]-\underset{h\to 0}{lim}\frac{b}{h}\left[\frac{x^2-{(x+h)}^2}{{(x+h)}^2·x^2}\right]+$

$\underset{h\to 0}{lim}\frac{1}{h}\left[-2sin\left(\frac{x+h+x}{2}\right)sin\left(\frac{x+h-x}{2}\right)\right]$

$=\underset{h\to 0}{lim}\frac{a}{h}\left[\frac{x^4-x^4-4x^{3}h-6x^2{h}^2-4x{h}^3-h^3}{x^4·{(x+h)}^4}\right]-\underset{h\to 0}{lim}\frac{b}{h}\left[\frac{x^2-x^2-h^2-2xh}{x^2(x+h){}^2}\right]$

$+\underset{h\to 0}{lim}\frac{a}{h}\left[-2sin\left(\frac{2x+h}{2}\right)sin\frac{h}{2}\right]$

$=\underset{h\to 0}{lim}\frac{a}{h}\left[\frac{h\left(-4x^3-6x^{2}h-4x{h}^2-h^2\right)}{x^4(x+h){}^4}\right]-\underset{h\to 0}{lim}\frac{b}{h}\left[\frac{h(-h-2x)}{x^2(x+h){}^2}\right]$

$-\underset{h\to 0}{lim}sin\left(\frac{2x+h}{2}\right)\underset{h\to 0}{lim}\frac{sin\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$h$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$=\underset{h\to 0}{lim}\frac{a\left(-4x^3-6x^{2}h-4x{h}^2-h^2\right)}{x^4(x+h){}^4}-\underset{h\to 0}{lim}\frac{b(-h-2x)}{x^2(x+h){}^2}$

$sin\left(\frac{2x+0}{2}\right)*1$

$=\frac{a\left(-4x^3\right)}{x^4·x^4}-\frac{b(-2x)}{x^2·x^2}-sinx.$

$=-\frac{4a}{x^5}+\frac{2b}{x^3}-sinx$

52. Given, f (x) = $\frac{p{x}^2+qx+r}{ax+b}$

$f^{\prime }(x)=\frac{(ax+b)\frac{d}{dx}\left(p{x}^2+qx+r\right)-\left(p{x}^2+qx+r\right)\frac{d}{dx}(ax+b)}{{(ax+b)}^2}$

$=\frac{(ax+b)(2xp+q)-\left(p{x}^2+qx+r\right)(a)}{{(ax+b)}^2}$

$=\frac{2ap{x}^2+2bpx+aqx+bq-ap{x}^2-aqx-ar}{{(ax+b)}^2}$

$=\frac{ap{x}^2+2bpx+bq-ar}{{(ax+b)}^2}$

51. Given, f (x) = $\frac{(ax+b)}{p{x}^2+qx+r}$

So, f(x) = $\frac{\left(p{x}^2+qx+r\right)\frac{d}{dx}(ax+b)-(ax+b)\frac{d}{dx}\left(p{x}^2+qx+r\right)}{{\left(p{x}^2+qx+r\right)}^2}$

 $=\frac{\left(p{x}^2+qx+r\right)a-(ax+b)(2xp+q)}{{\left(p{x}^2+qx+r\right)}^2}$

$=\frac{ap{x}^2+aqx+ar-2ap{x}^2-aqx-2bpx-b}{{\left(p{x}^2+qx+9\right)}^2}q.$

$=\frac{-ap{x}^2-2bpx+ar-bq}{{\left(p{x}^2+qx+r\right)}^2}$

48. Given, f (x) = $\frac{1}{a{x}^2+bx+c}$

So, f? (x) = $\frac{\left(a{x}^2+bx+c\right)\frac{d}{dx}(1)-1·\frac{d}{dx}\left(a{x}^2+bx+c\right)}{{\left(a{x}^2+bx+c\right)}^2}$

$=\frac{\left(a{x}^2+bx+c\right)(0)-(2ax+b)}{{\left(a{x}^2+bx+c\right)}^2}$

$=-\frac{(2ax+b)}{{\left(a{x}^2+bx+c\right)}^2}$

49. Given, f (x) = $\frac{1+\frac{1}{x}}{1-\frac{1}{x}}=\frac{\frac{x+1}{x}}{\frac{x-1}{x}}=\frac{x+1}{x-1}$

So, f (x) = $\frac{(x-1)\frac{d}{dx}(x+1)-(x+1)\frac{d}{dx}(x-1)}{{(x-1)}^2}$

= $\frac{(x-1)-(x+1)}{{(x-1)}^2}$

$=\frac{x-1-x-1}{{(x-1)}^2}=\frac{-2}{{(x-1)}^2}$