Question

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

Answer 1

`y' = cosxln(tanx) + secx`

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 `y`. The derivative can be found by using the product and chain rules.

Let `y = g(x)h(x)`, with `g(x) = sinx` and `h(x) = ln(tanx)`. The derivative of `g(x)`, by first principles, in `g'(x) = cosx`.

However, we must use the chain rule to find `h'(x)`. We let `y = lnu` and `u = tanx`. Then `dy/(du) = 1/u` and `(du)/dx = sec^2x`. This means that `h'(x) = 1/u * sec^2x = sec^2x/tanx = (1/cos^2x)/(sinx/cosx) = secxcscx`

We now apply the product rule, which states that `y' = g'(x)h(x) + h'(x)g(x)`.

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

`y' = cosxln(tanx) + secx`

Hopefully this helps!