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Question
if y = (x^3 - cos x)^5 then y =
a 5(x^3 - cos x)^4
b 5(3x^2 + sin x)^4
c 5(3x^2 + sin x)
d 5(3x^2 + sin x)^4cdot(6x + cos x)
e 5(x^3 - cos x)^4cdot(3x^2 + sin x)
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 $\frac{dy}{dx}$
$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=5(x^{3}-\cos x)^{4}\cdot(3x^{2}+\sin x)$
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E. $5(x^{3}-\cos x)^{4}\cdot(3x^{2}+\sin x)$