Maths

72. Let y = x^x - 2 sin x

Putting u = x^x and v = 2 sin x.

So, y = u - v

= $\frac{dy}{dx}=\frac{dy}{dx}-\frac{dv}{dx}$ ____ (i)

As u = x^x

Log u = x log x.

So,$\frac{d}{dx}$ log u = $\frac{d}{dx}$ x log x.

= $\frac{1}{4}\frac{dy}{dx}=\frac{xdlogx}{dx}+logx\frac{dx}{dx}$

x´ $\frac{1}{x}$ + log x.

= 1 + log x

= $\frac{dy}{dx}$= 4 [1 + log x] = x^x [1 + log x].

And v = 2sin x Log v = sin x log 2.

$\frac{d}{dx}(logv)=\frac{d}{dx}$(sin x log 2)

$\Rightarrow \frac{1}{v}\frac{dv}{dx}=sinx\frac{d}{dx}$ sin x $\frac{d}{dx}$ log 2 + log 2 $\frac{dsinx}{dx}$ = log 2. cos x.

$\Rightarrow \frac{dv}{dx}$ = v log2 cos x.

$\Rightarrow \frac{dv}{dx}$ = v log 2 cos x

= 2 ^{sin x} log². cos x.

Q Eqⁿ (i) becomes,    = x^x (1 + log x) - 2 ^{sin x} cos x log 2.

71. Let y = (log x) cos x

Taking loge on both sides,

Log y = cos x [log (log x)]

Differentiating w r t 'x' we get,

$\frac{1}{y}\frac{dy}{dx}=cosx\frac{d}{dx}$ log (log x) +log (log x) $\frac{dcosx}{dx}$

$=cosx*\frac{1}{logx}*\frac{d(logx)}{dx}$ + log (log x) (- sin x)

$=\frac{cogx}{xlogx}$ - sin x log (log x)

= $\frac{dy}{dx}=y{\left[\frac{cosx}{xlog2}-sinx·log(logx)\right]}_{.}$

$\Rightarrow \frac{dy}{dx}={(logx)}^{cosx}\left[\frac{cosx}{xlogx}-sinx·log(logx)\right]$

70. Kindly go through the solution

Putting value of y from the above we get,

69. Let y = cos x cos 2x cos 3x _____ (i)

Taking log_e on bolk sides.

logy = log (cos x) + log (cos 2x + log (cos 3x)

= log (cos x) + log (cos 2x) + log (cos 3x)

Differentiating w r t 'x'

$\frac{1}{y}\frac{dy}{dx}=\frac{1}{cosx}\frac{d}{dx}cosx+\frac{1}{cos2x}\frac{d}{dx}cos2x+\frac{1}{cos3x}\frac{d}{dx}cos3x.$

$=\frac{(-sinx)}{cosx}+\frac{(-sin2x)}{cos2x}\frac{d}{dx}(2x)+\frac{(-sin3x)}{cos3x}\frac{d(3x)}{dx}$

= - tan x-2 tan 2x- 3 tan 3x.

$\Rightarrow \frac{dy}{dx}$ = y [- tan x- 2 tan 2x- 3 tan 3x]

Putting value of y from (i) we get,

$=\frac{dy}{dx}$= - cos x cos 2x cos 3x [tan x + 2 tan 2x + 3 tan 3x]

68. Let y = cos (log x + e^x)

$\therefore \frac{dy}{dx}=\frac{d}{dx}$cos (log x + e^x)

= - sin (log x + e^x) $\frac{d}{dx}$ (log x + e^x)

= - sin (log x + e^x)  $\left[\frac{1}{x}+e^x\right]$

$=-\left[\frac{1}{x}+e^x\right]$ sin (log x + e^x).

67. $y=\frac{cotx}{logx}$

$\therefore \frac{dy}{dx}=\frac{logx\frac{d}{da}cox-cosx\frac{d}{dx}logx}{{[logx]}^2}$

$=\frac{-sinx·logx-cosx*\frac{1}{x}}{{[logx]}^2.}$

$=\frac{-[xsinxlogx-cosx]}{x{(logx)}^2.}$

 66. Let y = log (log x)

$\therefore \frac{dy}{dx}=\frac{1}{logx}\frac{d}{dx}(logx)$

$=\frac{1}{logx}*\frac{1}{x}$

$\frac{dy}{dx}=\frac{1}{xlogx}$

65. Kindly go through the solution

64. Let y = e^x + e^x2 + … + e^x5.

$\therefore \frac{dy}{dx}=\frac{d}{dx}$ (e^x + e^x2 + e^x3 + e^x4 + e^x5).

$=\frac{d{e}^x}{dx}+\frac{d{e}^{x^2}}{dx}+\frac{d{e}^{x^3}}{dx}+\frac{d}{dx}{e}^{x^4}+\frac{d}{dx}{e}^{x^{5.}}$

$=e^x+e^{x^2}\frac{d{x}^2}{dx}+e^{x^3}\frac{d{x}^3}{dx}+e^{x^4}\frac{d{x}^4}{dx}+e^{x^5}\frac{d{x}^5}{dx}.$

= e^x + 2x e^x2 + 3x²e^x3 + 4x³e^x4 + 5ee^x4

63. Let y = log (cos e^x).

$\therefore \frac{dy}{dx}=\frac{d}{d\mathrm{x }}$log (cos e^x)

$=\frac{1}{cos{e}^x}\frac{d}{dx}$ (cos e^x)

= - $\frac{}{}$$\frac{sin{e}^x}{cos{e}^x}\frac{d{e}^{x.}}{dx}$

= -e^x [tan e^x]