What is the derivative of f(t) = sin^2 (e^(sin^2t))?

Jan 24, 2017

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

Explanation:

We will peel off functions one by one using the chain rule. The overriding function is the squared function.

The chain rule states that the derivative of $g {\left(t\right)}^{2} = 2 g {\left(t\right)}^{1} \cdot g ' \left(t\right)$.

So, where $g \left(t\right) = \sin \left({e}^{{\sin}^{2} t}\right)$:

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

For the leftover derivative, the overriding function is the sine function. Where $\sin \left(h \left(t\right)\right)$, the derivative is $\cos \left(h \left(t\right)\right) \cdot h ' \left(t\right)$.

Where $h \left(t\right) = {e}^{{\sin}^{2} t}$, this becomes:

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

Now the overriding function is the power of $e$. For a function ${e}^{j \left(t\right)}$, the derivative through the chain rule is ${e}^{j \left(t\right)} \cdot j ' \left(t\right)$.

Here, $j \left(t\right) = {\sin}^{2} t$, so:

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

We again have a squared function. Following the method before:

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

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

Using $2 \sin \theta \cos \theta = \sin 2 \theta$ this can be rewritten as:

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