# How do you differentiate #e^y = (tanx)^sinx#?

##### 1 Answer

#### Explanation:

Take the natural logarithm of both sides.

#ln(e^y) = ln(tanx)^sinx#

#ylne = sinxln(tanx)#

#y = sinxln(tanx)#

This is now a normal function explicitly defined by

Let

However, we must use the chain rule to find

We now apply the product rule, which states that

#y' = cosx(ln(tanx)) + sinx(secxcscx)#

#y' = cosxln(tanx) + secx#

Hopefully this helps!