Maths

64.  Given, f (x) = $\frac{sin(x+a)}{cosx}$

$f^{\prime }(x)=\frac{cosx\frac{d}{dx}sin(x+a)-sin(x+a)\frac{d}{dx}cosx}{co{s}^{2}x}$

Let g?(x) = sin (x + a)

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

= cos (x + a)

And P(x) = cos x

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

Thus, f?(x) = $\frac{}{}$$\frac{cosx·cos(x+a)-sin(x+a)(-sinx)}{co{s}^{2}x}$

$=\frac{cosx·cos(x+a)+sin(x+a)sinx}{co{s}^{2}x}$

$=\frac{cos(x+a-x)}{co{s}^{2}x}$

$=\frac{cosa}{co{s}^{2}x}$

63. Given, f (x) = $\frac{a+bsinx}{c+dcosx}$

f?(x) = $\frac{(c+dcosx)\frac{d}{dx}(bsinx)-(a+bsinx)\frac{d}{dx}(c+dcosx)}{{(c+dcosx)}^2}$

$f^{\prime }(x)=\frac{(c+dcosx)·b·\frac{d}{dx}(sinx)-(a+bsinx)·d·\frac{d}{dx}(cosx)}{{(c+dcosx)}^2}\_\_\_\_\_(1)$

{Copy (A)}

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

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

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

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

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

= sin x ______ (2)

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

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

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

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

= cos x _____ (3)

So, put (2) and (3) in (1) we get,

$f^{\prime }(x)=\frac{(c+dcosx)(b·cosx)-(a+bsinx)(-d·sinx)}{{(c+dcosx)}^2}$

$=\frac{beccosx+bdco{s}^{2}x+adsinx+bdsi{n}^{2}x}{{(c+dcosx)}^2}$

$=\frac{bccosx+adsinx+bd\left(co{s}^{2}x+si{n}^{2}x\right)}{{(c+dcosx)}^2}$

$=\frac{bccosx+adsinx+bd}{{(c+dcosx)}^2}$

62. Given, f (x) =sinⁿx

By chain rule,

f? (x) = n (sin x)^n-1 $\frac{d}{dh}$ sin x

Let (gx) = sinx

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

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

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

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

= $cos\frac{(2x+0)}{2}*1$

= cos x.

So, f? (x) = n (sin x)^n-1 cos x.

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

$=\frac{\frac{1}{cosx}-1}{\frac{1}{cosx}+1}$

$=\frac{1-cosx}{1+cosx}$

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

$=\frac{(1+cosx)(-1)\frac{d}{dx}(-cosx)-(1-cosx)\frac{d}{dx}(cosx)}{{(1+cosx)}^2}$

Let g(x) = cos x.

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

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

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

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

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

= -sin x.

So, f?(x) $\frac{(1+cosx)(-1)(-sinx)-(1-cosx)(-sinx)}{{(1+cosx)}^2}$

$=\frac{sinx+cosxsinx+sinx-sinxcosx}{{(1+cosx)}^2}$

$=\frac{2sinx}{{(1+cosx)}^2}$

$=\frac{2sinx}{{\left(1+\frac{1}{secx}\right)}^2}=\frac{2sinx*se{c}^{2}x}{{(secx+1)}^2}=\frac{2secx·sinx}{{(sinx+1)}^2·cosx}.$

$=\frac{2secx·tanx}{{(sinx+1)}^2}$

60. Given, f (x) = $\frac{sinx+cosx}{sinx-cosx}$

So, f?(x) = $\frac{sinx-cosx\frac{d}{dx}\left(sinx+cosx\right)-\left(sinx+cos2\right)\frac{d}{dx}\left(sinx-cosx\right)}{{\left(sinx-cosx\right)}^2}$

Let g(x) = cos x and p(x) = sin x.

{from so g'(x) A ) (upto equation 3)

Let g(x) = cos2 and p(x) = sin x.

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

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

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

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

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

= -sin x ______ (2)

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

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

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

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

= cos x _____ (3)

Putting (2) and (3) in (1) we get,

$f^{\prime }(x)=\frac{(sinx-cosx)[cosx-sinx]-(csinx+cosx)[cosx+sinx]}{{(sinx-cosx)}^2}$

$=\frac{-{(sinx-cosx)}^2-{(sinx+cosx)}^2}{{(sinx-cosx)}^2}$

$=\frac{-\left(si{n}^{2}x+co{s}^{2}x\right)+2sinxcosx-\left(si{n}^{2}x+co{s}^{2}x\right)-2sinxcosx}{{(sinx-cosx)}^2}$

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

59. Given, f (x) = $\frac{cosx}{1+sinx}$

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

Putting (2) and (3) in (1) we get,

$f^{\prime }(x)=\frac{(1+sinx)(-sinx)-cosx(cosx)}{{(1+sinx)}^2}$

$=\frac{sinx-si{n}^{2}x-co{s}^{2}x}{{(1+sinx)}^2}$

$=\frac{-sinx-\left(si{n}^{2}x+co{s}^{2}x\right)}{{(1+sinx)}^2}$

$=\frac{-(sinx+1)}{{(1+sinx)}^2}=-\frac{1}{1+sinx}.$

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

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

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

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

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

Kindly go through the solution

Kindly go through the solution

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

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