Question

If
$y = {sin \: x}^{cos \: x}$
, find dy/dx.​

Answer 1

Answer:

Step-by-step explanation:

given y=$sinx^{cosx}$

by log on both sides we get

logy=cosx *logsinx

now diffrentiate with respect to x we get

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

dy/dx= (cosx-$sin^{2} x$logx)$sinx^{cosx}$/sinx

hope i answered the question