Continuity and Differentiability

81. Given, yx = xy

Taking log,

x log y .log x

Differentiating w r t 'x' we get,

$\Rightarrow x\frac{d}{dx}logy+logy\frac{dx}{dx}=y\frac{d}{dx}logx+logx\frac{dy}{dx}$

$\Rightarrow \frac{x}{y}·\frac{dy}{dx}+logy=\frac{y}{x}+logx\frac{dy}{dx}$

$\Rightarrow logx\frac{dy}{dx}-\frac{x}{y}\frac{dy}{dx}=logy-\frac{y}{x}$

$\Rightarrow \frac{dy}{dx}\left[\frac{ylogx-x}{y}\right]=\frac{xlogy-y}{x}$

$\Rightarrow \frac{dy}{dx}=\frac{y(xlogy-y)}{x(ylogx-x)}.$

80. Given, xy + yx = 1

Let 4 = xy and v =., we have,

u + v = 1.

$\Rightarrow \frac{dy}{dx}+\frac{dv}{dy}=0$ ___ (1)

So, u = xy

= log u = y log x(taking log)

Now, differentiating w r t 'x',

$\frac{1}{4}\frac{dy}{dx}=y\frac{d}{dx}logx+logx\frac{dy}{dx}.$

$\frac{dy}{dx}=4\left[\frac{y}{x}+logx\frac{dy}{dx}\right]$

$\Rightarrow x^y·\frac{y}{x}+x^{y}logx·\frac{dy}{dx}$

= xy- 1y + xy log x $\frac{dy}{dx}.$

And v = yx.

log v = x log y.

Differentiating w r t 'x',

$\Rightarrow \frac{1}{v}\frac{dv}{dx}=x\frac{d}{dx}logy+logy\frac{dx}{dx}$

$=\frac{x}{y}\frac{dy}{dx}+logy$

$\Rightarrow \frac{dv}{dx}=v\left[\frac{x}{y}\frac{dy}{dx}+logy\right]$

$=y^x·\frac{x}{y}·\frac{dy}{dx}+y^{x}log·y$

= yx- 1. $x\frac{dy}{dx}$ + yx log y.

So, eqn (1) becomes

xy- 1y + xy log x $\frac{dy}{dx}$ + yx - 1 $\frac{dy}{dx}$ + yx log y = 0

$\Rightarrow \frac{dy}{dx}\left(x^{y}logx+y^{x+1}·x\right)$ = - (xy- 1y + yx log y)

$\Rightarrow \frac{dy}{dx}=\frac{-\left(x^{y-1}·y+y^{x}logy\right)}{\left(x^{y}logx+y^{x-1}·x\right).}$

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,

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