Question

If log( log y) =xlog x + log( log x)then the value of dy/dx is
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Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\: log(logy) = xlogx + log(logx)$

Now, Differentiate both sides with respect to x,

$\rm :\longmapsto\: \dfrac{d}{dx} log(logy) =\dfrac{d}{dx}( xlogx + log(logx) )$

We know that

$\boxed{ \sf{ \: \dfrac{d}{dx}(u + v) = \dfrac{d}{dx}u + \dfrac{d}{dx}v}}$

and

$\boxed{ \sf{ \: \dfrac{d}{dx}logx = \frac{1}{x}}}$

$\rm :\longmapsto\: \dfrac{1}{logy} \dfrac{d}{dx}(logy) =\dfrac{d}{dx}xlogx + \dfrac{d}{dx} log(logx)$

We know that

$\boxed{ \sf{ \: \dfrac{d}{dx}(uv) = u\dfrac{d}{dx}v + v\dfrac{d}{dx}u}}$

$\rm :\longmapsto\: \dfrac{1}{logy} \dfrac{1}{y} \dfrac{dy}{dx}=x\dfrac{d}{dx}logx + logx\dfrac{d}{dx}x+ \dfrac{1}{logx} \dfrac{d}{dx}(logx)$

$\rm :\longmapsto\: \dfrac{1}{logy} \dfrac{1}{y} \dfrac{dy}{dx}=x \times \dfrac{1}{x} + logx \dfrac{1}{logx} \times \dfrac{1}{x}$

$\rm :\longmapsto\: \dfrac{1}{ylogy} \dfrac{dy}{dx}=1 + logx + \dfrac{1}{xlogx}$

$\rm :\longmapsto\:\dfrac{dy}{dx}=y \: logy \: \bigg(1 + logx + \dfrac{1}{xlogx} \bigg)$

More to know :-

$\boxed{ \sf{ \: \dfrac{d}{dx}x = 1}}$

$\boxed{ \sf{ \: \dfrac{d}{dx}k = 0}}$

$\boxed{ \sf{ \: \dfrac{d}{dx} {e}^{x} = {e}^{x} }}$

$\boxed{ \sf{ \: \dfrac{d}{dx} {a}^{x} = {a}^{x} log(a) }}$