If y= 3^ (sin x), what is dy/dx?

It is given that y = 3^ (sin
x)

We take the natural logarithm of both the sides and use
the relation ln a^b = b*ln a.

=> ln y = ln (3^sin
x)

=> ln y = (sin x) (ln
3)

Now differentiate both the sides with respect to
x

=> (1/y) (dy/dx) = (ln 3) cos
x

=> dy/dx = (ln 3)*y*cos
x

=> dy /dx = 3^ (sin x)*(cos x)* ln
3

Therefore if y= 3^ (sin x), dy /dx = 3^
(sin x)*(cos x)* ln 3.