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

Dec 20, 2015

$f ' \left(t\right) = 2 \sin \left({e}^{{\sin}^{2} t}\right) \cdot \cos \left({e}^{{\sin}^{2} t}\right) \cdot {e}^{{\sin}^{2} t} \cdot 2 \sin t \cos t$

#### Explanation:

We will need to use all of the following rules of differentiation :

$\frac{d}{\mathrm{dx}} \sin \left(u \left(x\right)\right) = \cos u \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{d}{\mathrm{dx}} {e}^{u \left(x\right)} = {e}^{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{d}{\mathrm{dx}} {\left(u \left(x\right)\right)}^{n} = n {\left(u \left(x\right)\right)}^{n - 1} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Applying all these rules in conjunction yields \

$f ' \left(t\right) = 2 \sin \left({e}^{{\sin}^{2} t}\right) \cdot \cos \left({e}^{{\sin}^{2} t}\right) \cdot {e}^{{\sin}^{2} t} \cdot 2 \sin t \cos t$