Question

find the derivative of the function.

$f(t)=sin^{2}(e^{sin^{2}(t)})$

$f(t)=$

Explanation:

Step1: Apply chain - rule for outer function

Let $u = e^{\sin^{2}(t)}$, then $f(t)=\sin^{2}(u)$. The derivative of $y = \sin^{2}(u)$ with respect to $u$ is $y^\prime=2\sin(u)\cos(u)$ by the power - rule and the derivative of $\sin(u)$.

Step2: Apply chain - rule for middle function

Now we need to find the derivative of $u = e^{\sin^{2}(t)}$ with respect to $t$. Let $v=\sin^{2}(t)$. Then $u = e^{v}$, and the derivative of $u$ with respect to $v$ is $\frac{du}{dv}=e^{v}$.

Step3: Apply chain - rule for inner function

Next, find the derivative of $v=\sin^{2}(t)$ with respect to $t$. Let $w = \sin(t)$, then $v = w^{2}$. The derivative of $v$ with respect to $w$ is $2w$, and the derivative of $w=\sin(t)$ with respect to $t$ is $\cos(t)$. So $\frac{dv}{dt}=2\sin(t)\cos(t)$.

Step4: Combine using chain - rule

By the chain - rule $\frac{df}{dt}=\frac{df}{du}\cdot\frac{du}{dv}\cdot\frac{dv}{dt}$.
Substitute back $u = e^{\sin^{2}(t)}$ and $v=\sin^{2}(t)$:
\[

$$\begin{align*} f^\prime(t)&=2\sin(e^{\sin^{2}(t)})\cos(e^{\sin^{2}(t)})\cdot e^{\sin^{2}(t)}\cdot2\sin(t)\cos(t)\\ &=2\sin(t)\cos(t)\cdot2\sin(e^{\sin^{2}(t)})\cos(e^{\sin^{2}(t)})\cdot e^{\sin^{2}(t)} \end{align*}$$

\]

Answer:

$2\sin(t)\cos(t)\cdot2\sin(e^{\sin^{2}(t)})\cos(e^{\sin^{2}(t)})\cdot e^{\sin^{2}(t)}$