Question

f(x)=\cos\left(e^{\left(\sqrt{x^{2}+x}\
ight)}\
ight) **challenge 4 chains!

Explanation:

Step1: Let $u = e^{\sqrt{x^{2}+x}}$.

Then $f(x)=\cos(u)$.

Step2: Differentiate $y = \cos(u)$ with respect to $u$.

$\frac{dy}{du}=-\sin(u)$

Step3: Now let $v=\sqrt{x^{2}+x}=(x^{2}+x)^{\frac{1}{2}}$.

Since $u = e^{v}$, differentiate $u$ with respect to $v$.
$\frac{du}{dv}=e^{v}$

Step4: Differentiate $v=(x^{2}+x)^{\frac{1}{2}}$ with respect to $x$.

Using the chain - rule and power - rule, $\frac{dv}{dx}=\frac{1}{2}(x^{2}+x)^{-\frac{1}{2}}(2x + 1)=\frac{2x + 1}{2\sqrt{x^{2}+x}}$

Step5: By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dv}\cdot\frac{dv}{dx}$.

Substitute the values of $\frac{dy}{du}$, $\frac{du}{dv}$ and $\frac{dv}{dx}$:
$\frac{dy}{dx}=-\sin(e^{\sqrt{x^{2}+x}})\cdot e^{\sqrt{x^{2}+x}}\cdot\frac{2x + 1}{2\sqrt{x^{2}+x}}$

Answer:

$-\frac{(2x + 1)e^{\sqrt{x^{2}+x}}\sin(e^{\sqrt{x^{2}+x}})}{2\sqrt{x^{2}+x}}$