if y = (x^3 - cos x)^5 then y =\na 5(x^3 - cos x)^4\nb 5(3x^2 + sin x)^4\nc 5(3x^2 + sin x)\nd 5(3x^2 + sin…

Explanation:

Step1: Apply chain - rule

Let $u = x^{3}-\cos x$, then $y = u^{5}$. The chain - rule states that $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$.

Step2: Differentiate $y$ with respect to $u$

$\frac{dy}{du}=5u^{4}=5(x^{3}-\cos x)^{4}$

Step3: Differentiate $u$ with respect to $x$

$\frac{du}{dx}=\frac{d}{dx}(x^{3}-\cos x)=3x^{2}+\sin x$

Step4: Calculate $\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=5(x^{3}-\cos x)^{4}\cdot(3x^{2}+\sin x)$

Answer:

E. $5(x^{3}-\cos x)^{4}\cdot(3x^{2}+\sin x)$