Question

if f(x)=e^sin²x, then f(x)= o e^x sin²x o e^sin x(sin x + 2 cos x) o 2e^x sin x cos x o 2e^sin²x sin x cos x question 3 of 26

Explanation:

Step1: Apply chain - rule

Let $u = \sin^{2}x$. Then $f(x)=e^{u}$. The derivative of $y = e^{u}$ with respect to $u$ is $\frac{dy}{du}=e^{u}$, and we need to find $\frac{du}{dx}$.

Step2: Differentiate $u=\sin^{2}x$

Using the chain - rule again, if $u = v^{2}$ where $v=\sin x$. Then $\frac{du}{dv} = 2v$ and $\frac{dv}{dx}=\cos x$. So $\frac{du}{dx}=\frac{du}{dv}\cdot\frac{dv}{dx}=2\sin x\cos x$.

Step3: Find $f'(x)$

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substituting $u = \sin^{2}x$ and $\frac{du}{dx}=2\sin x\cos x$ and $\frac{dy}{du}=e^{u}$, we get $f'(x)=e^{\sin^{2}x}\cdot2\sin x\cos x = 2e^{\sin^{2}x}\sin x\cos x$.

Answer:

$2e^{\sin^{2}x}\sin x\cos x$