Question

find the derivative of the given function.
y = x² sin³x + x cos⁻⁵x
\frac{dy}{dx} = \square

Explanation:

Step1: Apply sum rule

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

Step2: Apply product rule on first term

Product rule: $(uv)^\prime = u^\prime v+uv^\prime$. Let $u = x^{2}$, $v=\sin^{3}x$.
$u^\prime=2x$. For $v=\sin^{3}x$, use chain rule: $v^\prime = 3\sin^{2}x\cos x$.
$$\frac{d}{dx}(x^{2}\sin^{3}x)=2x\sin^{3}x+x^{2}\cdot3\sin^{2}x\cos x=2x\sin^{3}x + 3x^{2}\sin^{2}x\cos x$$

Step3: Apply product rule on second term

Let $u = x$, $v=\cos^{-5}x$. $u^\prime = 1$. For $v=\cos^{-5}x$, use chain rule: $v^\prime=- 5\cos^{-6}x(-\sin x)=5\cos^{-6}x\sin x$.
$$\frac{d}{dx}(x\cos^{-5}x)=\cos^{-5}x+x\cdot5\cos^{-6}x\sin x=\cos^{-5}x + 5x\cos^{-6}x\sin x$$

Step4: Combine results

$$\frac{dy}{dx}=2x\sin^{3}x+3x^{2}\sin^{2}x\cos x+\cos^{-5}x + 5x\cos^{-6}x\sin x$$

Answer:

$$2x\sin^{3}x+3x^{2}\sin^{2}x\cos x+\cos^{-5}x + 5x\cos^{-6}x\sin x$$