Question

If
$y = {sinx}^{ {sinx}^{sinx.... \infty } }$
prove that

$\frac{dy}{dx} = \frac{ {y}^{2}cotx }{1 - ylogsinx}$

​

Answer 1

$\huge\red{\color{pink}{\fcolorbox{cyan}{black}{solution}}}$

Given:

$\red{ \star} y = { sinx}^{ {sinx}^{sinx... \infty } }$

To prove:

$\red\star \: \frac{dy}{dx} = \frac{ {y}^{ ^{2} } cotx}{ 1 - y log \: sinx}$

$\red{\color{white}{\fcolorbox{cyan}{black}{Answer:-}}}$

$\leadsto \: y = {sinx}^{ {sinx}^{sinx.. \infty } } \\ \\ \leadsto \: y = ({sinx})^{y}$

• Take Log both sides , we get

$\leadsto \: log(y) = log(( {sinx})^{y} ) \\ \\ \leadsto \: log(y) = y log(sinx)$

•Differentiate both sides w.r.t X

$\leadsto \: \frac{d}{dx} log(y) = \frac{d}{dx} (y \: log(sinx) ) \ \\ \\ \leadsto \: \frac{1}{y} \frac{dy}{dx} = y \frac{d}{dx} ( log(sinx) ) + log(sinx) \frac{dy}{dx} \\ \\ \leadsto \: \frac{1}{y} \frac{dy}{dx} = \frac{y}{sinx} ( cosx) + log(sinx) \frac{dy}{dx} \\ \\ \leadsto \: \frac{dy}{dx} ( \frac{1}{y} - log(sinx) ) = y( cotx) \\ \\ \leadsto \: \frac{dy}{dx} ( \frac{1 - y log(sinx) }{y} ) = y( cotx) \\ \\ \leadsto \: \frac{dy}{dx} = \frac{ {y}^{2}cotx }{1 - y log(sinx) }$

$\huge\red{\color{green}{\fcolorbox{cyan}{black}{proved}}}$

$\red{\color{white}{\fcolorbox{cyan}{black}{formula}}}$

$\star \bold{ \frac{d}{dx} (u.v) = u \frac{dv}{dx} + v \frac{du}{dx} }$

$\star \bold{ \frac{d}{dx} log(x) = \frac{1}{x} } \\ \\ \star \bold{ \frac{d}{dx} sinx = cosx}$

Answer 2

Formulae used:

$\frac{d}{dx} (x.y) = x \frac{d}{dx} (y) + y \frac{d}{dx} (x) \\ \\ \frac{d}{dx} (sinx) = cosx \\ \\ \frac{d}{dx} (logy) = \frac{1}{y}$