# How do you find the derivative of  f(t)=sin^2[e^(sin^2)t] using the chain rule?

Oct 30, 2015

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

#### Explanation:

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

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

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

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

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

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