Question

f(x)=sqrt{cos(e^{x^{3}}sin(x))}
f(x)=?

Explanation:

Step1: Identify the outermost - function

The outermost function is the square - root function $y = \sqrt{u}=\ u^{\frac{1}{2}}$, where $u=\cos(e^{x^{3}}\sin(x))$. By the power - rule and chain - rule, $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$.

Step2: Identify the middle - function

The middle - function is $u = \cos(v)$, where $v = e^{x^{3}}\sin(x)$. By the derivative of cosine function and chain - rule, $\frac{du}{dv}=-\sin(v)$.

Step3: Identify the innermost - function

The innermost function is $v = e^{x^{3}}\sin(x)$. Use the product rule $(ab)^\prime=a^\prime b + ab^\prime$, where $a = e^{x^{3}}$ and $b=\sin(x)$.

For $a = e^{x^{3}}$

Let $t=x^{3}$, then $a = e^{t}$. By the chain - rule, $\frac{da}{dt}=e^{t}$ and $\frac{dt}{dx}=3x^{2}$, so $\frac{da}{dx}=e^{x^{3}}\cdot3x^{2}$.

For $b=\sin(x)$

$\frac{db}{dx}=\cos(x)$.
Then $\frac{dv}{dx}=e^{x^{3}}\cdot3x^{2}\sin(x)+e^{x^{3}}\cos(x)=e^{x^{3}}(3x^{2}\sin(x)+\cos(x))$.

Step4: Calculate $f^\prime(x)$

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dv}\cdot\frac{dv}{dx}$.
\[

$$\begin{align*} f^\prime(x)&=\frac{1}{2}(\cos(e^{x^{3}}\sin(x)))^{-\frac{1}{2}}\cdot(-\sin(e^{x^{3}}\sin(x)))\cdot e^{x^{3}}(3x^{2}\sin(x)+\cos(x))\\ &=-\frac{e^{x^{3}}(3x^{2}\sin(x)+\cos(x))\sin(e^{x^{3}}\sin(x))}{2\sqrt{\cos(e^{x^{3}}\sin(x))}} \end{align*}$$

\]

Answer:

$-\frac{e^{x^{3}}(3x^{2}\sin(x)+\cos(x))\sin(e^{x^{3}}\sin(x))}{2\sqrt{\cos(e^{x^{3}}\sin(x))}}$