Question

How do you differentiate `f(t)=ln(e^(sin^2t))` using the chain rule?

Answer 1

Without rewriting

Use `d/dx lnu = 1/u (du)/dx` and `d/dx(e^u) = e^u (du)/dx` toi get

`f'(t) = 1/e^(sin^2t)[d/dt(e^(sin^2t))]`

`= 1/e^(sin^2t)[(e^(sin^2t)d/dt(sin^2t)]`

`= e^(sin^2t)/(e^(sin^2t))[2sintd/dx(sint))]`

`= 2sintcost = sin2t`

With rewriting

Use `ln(e^u) = u` to rewrite

`f(x) = sin^2x = (sinx)^2`

Now use the chain rule to get

`f'(x) = 2sint[d/dx(sint)] = 2sintcost = sin2t`