What is the derivative of #y = (sin x)^(cos x)#?

1 Answer
Jan 4, 2017

Answer:

#dy/dx = (sinx)^cosx(-sinxln(sinx) + cosxcotx)#

Explanation:

Take the natural logarithm of both sides.

#lny = ln(sinx)^cosx#

Use the rule #loga^n = nloga# to simplify:

#lny = cosxln(sinx)#

Use the implicit differentiation as well as the product and chain rules to differentiate.

#d/dx(lnsinx) = 1/sinx * cosx = cosx/sinx = cotx#

Now to the relation as a whole:

#1/y(dy/dx) = -sinx(ln(sinx)) + cosxcotx#

Solve for #dy/dx#:

#dy/dx = y(-sinxln(sinx) + cosxcotx)#

#dy/dx = (sinx)^cosx(-sinxln(sinx) + cosxcotx)#

Hopefully this helps!