Question

What will be the differentiation of sin x ki power sin x ki power sin x ki power sin x upto infinity

Answer 1

This is how its done

Answer 2

Given: $\mathsf{y={sinx}^{{sinx}^{{sinx}^{...\infty}}}}$

Taking log to both sides, we get

$\quad \mathsf{log(y)=log(sinx)^{y}}$

$\to \mathsf{log(y)=y\:log(sinx)}$

∴ differentiating both sides w. r. to x, we get

$\quad \mathsf{\frac{d}{dx}\{log(y)\}=\frac{d}{dx}\{y\:log(sinx)\}}$

$\to \mathsf{\frac{1}{y}\:\frac{dy}{dx}=y\:cotx+log(sinx)\:\frac{dy}{dx}}$

$\quad\quad \mathsf{[\because \frac{d}{dx}\{log(sinx)\}=cotx]}$

$\to \mathsf{\big[\frac{1}{y}-log(sinx)\big]\:\frac{dy}{dx}=y\:cotx}$

$\to \mathsf{\frac{dy}{dx}=\frac{y\:cotx}{\frac{1}{y}-log(sinx)}}$

$\to \mathsf{\frac{dy}{dx}=\frac{y^{2}\:cotx}{1-y\:log(sinx)}}$

$\to \mathsf{\frac{d}{dx}\big[{sinx}^{{sinx}^{{sinx}^{...\infty}}}\big]=\frac{\big[{sinx}^{{sinx}^{{sinx}^{...\infty}}}\big]^{2}\:cotx}{1-{sinx}^{{sinx}^{{sinx}^{...\infty}}}\:log(sinx)}}$

This is the required differentiation.

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