Question

y=(sinx)^sinx,Find dy/dx for the given function y wherever defined

Answer 1

Given, $\bf{y=(sinx)^{sinx}}$

$\implies y=e^{sinx.log(sinx)}$

differentiate with respect to x,

$\frac{dy}{dx}=\frac{d\{e^{sinx.log(sinx)}\}}{dx}\\\\=e^{sinx.log(sinx)}.\frac{d\{sinx.log(sinx)\}}{dx}\\\\=e^{sinx.log(sinx)}.[sinx\frac{d\{log(sinx)\}}{dx}+log(sinx).\frac{d(sinx)}{dx}]\\\\=e^{sinx.log(sinx)}[sinx.\frac{1}{sinx}\frac{d(sinx)}{dx}+log(sinx).cosx]\\\\=(sinx)^{sinx}[cosx+log(sinx).cosx]$

hence, dy/dx = (sinx)^(sinx).cosx[1 + log(sinx)]

Answer 2

HELLO DEAR,



GIVEN:-
Y = (sinx)^(sinx)


talking log both of side,

log y = (sinx).log(sinx)

differentiating y w.r.t x

(1/y)*dy/dx = (sinx)*d{log(sinx)}/dx + log(sinx)*d(sinx)/dx

(1/y)*dy/dx = sinx*{cosx/sinx} + log(sinx) * cosx

(1/y)*dy/dx = cosx + cosx*log(sinx)

dy/dx = ( y )*cosx{1 + log(sinx)}

dy/dx = (sinx)^(sinx)*cosx{1 + log(sinx)]}


I HOPE ITS HELP YOU DEAR,
THANKS