Question

find the derivative of y^(siny)=x^(sinx)

Answer 1

Take the log of both sides and use implicit differentiation.
\[ \sin(y)\log(y) = \sin(x)\log(x) \]
\[\left( \cos(y)\log(y) + \frac{\sin(y)}{y} \right)dy = \left(\cos(x)\log(x) + \frac{\sin(x)}{x}\right)dx \]
so
\[\frac{dy}{dx} = \frac{\cos(x)\log{x} + \frac{\sin(x)}{x}}{\cos(y)\log(y) + \frac{\sin(y)}{y}} \]

Answer 2

thanx

Answer 3

(x^(-1+sin(x)) y^(1-sin(y)) (x cos(x) log(x)+sin(x)))/(y cos(y) log(y)+sin(y)) is it possible to get this as an answer

Answer 4

Yes, that's also fine. It's the same as what I have up there, except for a factor
\[ \frac{x^{\sin(x)}}{y^{\sin(y)}} \]
but of course that's okay, because
\[y^{\sin(y)} = x^{\sin(x)} \]
so that fraction just equals 1.