Ncert Solutions Maths class 12th

78. Let y = x^x cos x  $\frac{a^2+1}{x^2-1}$

Putting  4 = x^x cos x and v = $\frac{x^2+1}{x^2-1}$ we have,

y = u + v

$\Rightarrow \frac{dy}{dx}=\frac{dy}{dx}+\frac{dv}{dx}$ ____ (1)

As u $\left|=\right|$ x^x cos x.

Taking log,

Log u = x cos x log x

Differentiating w r t 'x',

$\frac{1}{u}\frac{dy}{dx}=x\frac{d}{dx}$ [cos x log x] + cos x log x $\frac{dx}{dx}$

= x $\left\{cotx\frac{d}{dx}logx+logx\frac{d}{dx}cosx\right\}$ + cos x log x.

$=x\left\{cosx·\frac{1}{x}-sinx·logx\right\}$ + cos x log x.

= cos x- sin x. log x + cos x log x.

$\frac{dy}{dx}=u$ [cosx + cos x log x- sin x log x]

= x^x cos x [cos x + cos x log x-x sin x log x]

And v = $\frac{x^2+1}{x^2-1}$

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

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

$=\frac{2x^3-2x-2x^3-2x}{{\left(x^2-1\right)}^2}$

$=\frac{-4x}{{\left(x^2-1\right)}^2}.$

Hence, eqn (1) becomes,

$\frac{dy}{dx}$ x^{xcos x} [cos x + cos x log x-x sin x log x] $\frac{-4x}{{\left(x^2-1\right)}^2}$

76. Kindly go through the solution

75. Let y = (log x)^x + x ^{log x}.

Putting u = log x^x and v = x log x we get,

y = u + v

$\Rightarrow \frac{dy}{dx}=\frac{dy}{dx}+\frac{dv}{dx}$ .____ (1)

As u = log x^x

Taking log,

Þlog u = x [log(log x)]

Differentiating w r t x we get,

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

= $x*\frac{1}{logx}\frac{dlogx}{dx}$ + log (log x)

= $\frac{x}{logx}*\frac{1}{x}+log1(logx)$

$\Rightarrow \frac{dy}{dx}=\mu \left[\frac{1}{logx}+log(logx)\right]$

$\frac{dy}{dx}={(logx)}^x{\left[\frac{1}{logx}+log(logx)\right]}_{.}$

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

= (log x)^{x- 1} [1 + log ´. log (log x)]

And v = log x

Taking log,

Log v = log x log x. = (log x)².

Differentiating w r t 'x' we get,

$\Rightarrow \frac{1}{v}\frac{dv}{dx}=2logx\frac{d}{dx}logx$

$\Rightarrow \frac{dv}{dx}$ = 2v log x$\frac{1}{x}$

= 2. x log x. $\frac{logx}{x}$

= 2 x log x- 1 log x.

Hence eqⁿ becomes

$\frac{dy}{dx}$= (log x) x- 1[1 + log x log (log x)] + 2x ^{log x- 1} log x

74 . Let y = ${\left(x+\frac{1}{x}\right)}^x$ + ${\left(1+\frac{1}{x}\right)}^x$

Putting u = ${\left(x+\frac{1}{x}\right)}^x$ and v = ${\left(1+\frac{1}{x}\right)}^x$ we get,

y = u + v

$\therefore \frac{dy}{dx}=\frac{du}{dx}+\frac{dv}{dx}$ _____ (1)

As u = ${\left(x+\frac{1}{x}\right)}^x$

Taking log,

= log u = x log $\left(x+\frac{1}{x}\right)$

Differentiating w r t 'x' we get,

$\frac{1}{4}\frac{dy}{dx}=x\frac{d}{dx}log$ $\left(x+\frac{1}{x}\right)+log\left(x+\frac{1}{x}\right)\frac{dx}{dx}$

$=x\frac{1}{\left(x+\frac{1}{x}\right)}\frac{d}{dx}\left(x+\frac{1}{x}\right)$ + log $\left(x+\frac{1}{x}\right)$ 1.

$=x·\frac{1}{\left(x+\frac{1}{x}\right)}*\left(1-\frac{1}{x^2}\right)+log\left(x+\frac{1}{x}\right)$

$=\frac{x.\frac{x^2-1}{x^2}}{\left(\frac{x^2+1}{x}\right)}+log\left(x+\frac{1}{x}\right).$

$=\frac{x^2-1}{x^2+1}+log\left(x+\frac{1}{x}\right).$

$\Rightarrow \frac{du}{dx}=u\left[\frac{x^2-1}{x^2+1}+log\left(x+\frac{1}{x}\right)\right]$

$={\left(x+\frac{1}{x}\right)}^x{\left[\frac{x^2-1}{x^2+1}+log\left(x+\frac{1}{x}\right)\right]}_{.}$

And v = x $\left(1+\frac{1}{x}\right)$

Taking log, log v = $\left(1+\frac{1}{x}\right)$ log x

Differentiating W r t 'x',

$\frac{1}{v}\frac{dv}{dx}=\left(1+\frac{1}{x}\right)\frac{d}{dx}$ log x + log x $\frac{d}{dx}{\left(1+\frac{1}{x}\right)}_{.}$

$\left(1+\frac{1}{x}\right)*\frac{1}{x}$ + log x ${\left(0-\frac{1}{x^2}\right)}_{.}$

$\frac{dv}{dx}$ = v $\left[\frac{1}{x}\left(1+\frac{1}{x}\right)-\frac{logx}{x^2}\right]$ = $x^{\left(1+\frac{1}{x}\right)}\left[\frac{1}{x}\left(1+\frac{1}{x}\right)-\frac{logx}{x^2}\right]$

Hence, eqn (1) becomes,

$\frac{dy}{dx}={\left(x+\frac{1}{x}\right)}^x{\left[\frac{x^2-1}{x^2+1}+log\left(x+\frac{1}{x}\right)\right]}_{.}$ $+x^{\left(1+\frac{1}{x}\right)}\left[\frac{1}{x}\left(1+\frac{1}{x}\right)-\frac{logx}{x^2}\right]$

73. Let y = (x + 3)² (x + 4)³ (x + 5)⁴.

Taking log_e on both sides,

log y = log (x + 3)² + log (x + 4)³ + log (x + 5)⁴

= 2 log (x + 3) + 3 (log (x + 4) + 4 log (x + 5).

So,

Q $\frac{d}{dx}$ log y = $\frac{d}{dx}$  [2 log (x + 3) + 3 log (x + 4) + 4 log (x +5)]

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5.}$

$\Rightarrow \frac{dy}{dx}=y\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right]$

$\Rightarrow \frac{dy}{dx}$ = (x + 3)² (x + 4)³ (x + 5)⁴${\left[\frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5}\right]}_{.}$

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]