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!