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

##### 2 Answers
Mar 16, 2016

$\frac{\mathrm{dy}}{\mathrm{dt}} = {\cos}^{2} t \left({e}^{{\sin}^{2}} t\right) \cdot {e}^{{\sin}^{2}} t \cdot {\cos}^{2} t$

#### Explanation:

So, we got three functions here:

${\sin}^{2} \left({e}^{{\sin}^{2}} t\right)$

${e}^{{\sin}^{2}} t$

and

Let $y = {\sin}^{2} \left({e}^{{\sin}^{2}} t\right)$

differentiating w.r.t. $t$

$\frac{\mathrm{dy}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} {\sin}^{2} \left({e}^{{\sin}^{2}} t\right)$

$\frac{\mathrm{dy}}{\mathrm{dt}} = {\cos}^{2} t \left({e}^{{\sin}^{2}} t\right) \cdot \frac{d}{\mathrm{dt}} {e}^{{\sin}^{2}} t$

$\frac{\mathrm{dy}}{\mathrm{dt}} = {\cos}^{2} t \left({e}^{{\sin}^{2}} t\right) \cdot {e}^{{\sin}^{2}} t \cdot \frac{d}{\mathrm{dt}} {\sin}^{2} t$

$\frac{\mathrm{dy}}{\mathrm{dt}} = {\cos}^{2} t \left({e}^{{\sin}^{2}} t\right) \cdot {e}^{{\sin}^{2}} t \cdot {\cos}^{2} t$

This will be the differentiated function.

Mar 16, 2016

Please see the explanation below.

#### Explanation:

$f \left(t\right) = {\sin}^{2} \left({e}^{{\sin}^{2} t}\right)$ has the form

${\sin}^{2} \left(u\right)$ which is also ${\left(\sin \left(u\right)\right)}^{2}$

So we need the derivative of a square and we'll need the chain rule.

$\frac{d}{\mathrm{dt}} \left({\left(\sin \left(u\right)\right)}^{2}\right) = 2 \left(\sin \left(u\right)\right) \cdot \frac{d}{\mathrm{dt}} \left(\sin \left(u\right)\right)$

$= 2 \sin \left(u\right) \cos \left(u\right) \frac{d}{\mathrm{dt}} \left(u\right)$

In this problem, $u = {e}^{{\sin}^{2} t}$

So,
$\frac{\mathrm{du}}{\mathrm{dt}} = {e}^{{\sin}^{2} t} \cdot \frac{d}{\mathrm{dt}} \left({\sin}^{2} t\right)$

$= {e}^{{\sin}^{2} t} \cdot \left[2 \sin t \cos t\right]$

Combining all of this into one calculation:

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

$= 4 \sin t \cos t \cdot {e}^{{\sin}^{2} t} \sin \left({e}^{{\sin}^{2} t}\right) \cos \left({e}^{{\sin}^{2} t}\right)$

Because $2 \sin \theta \cos \theta = \sin \left(2 \theta\right)$, this couls also be written as

$= \left[2 \sin t \cos t\right] \cdot \left[{e}^{{\sin}^{2} t}\right] \left[2 \sin \left({e}^{{\sin}^{2} t}\right) \cos \left({e}^{{\sin}^{2} t}\right)\right]$

$= \sin \left(2 t\right) {e}^{{\sin}^{2} t} \sin \left(2 {e}^{{\sin}^{2} t}\right)$