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

Apr 24, 2016

Without rewriting

Use $\frac{d}{\mathrm{dx}} \ln u = \frac{1}{u} \frac{\mathrm{du}}{\mathrm{dx}}$ and $\frac{d}{\mathrm{dx}} \left({e}^{u}\right) = {e}^{u} \frac{\mathrm{du}}{\mathrm{dx}}$ toi get

$f ' \left(t\right) = \frac{1}{e} ^ \left({\sin}^{2} t\right) \left[\frac{d}{\mathrm{dt}} \left({e}^{{\sin}^{2} t}\right)\right]$

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

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

$= 2 \sin t \cos t = \sin 2 t$

With rewriting

Use $\ln \left({e}^{u}\right) = u$ to rewrite

$f \left(x\right) = {\sin}^{2} x = {\left(\sin x\right)}^{2}$

Now use the chain rule to get

$f ' \left(x\right) = 2 \sin t \left[\frac{d}{\mathrm{dx}} \left(\sin t\right)\right] = 2 \sin t \cos t = \sin 2 t$