Question

find the derivative of the given function.
y = x² sin⁷x + x cos⁻²x
dy/dx =

Explanation:

Step1: Differentiate \(x^{2}\sin^{7}x\) using product rule \((uv)^\prime = u^\prime v+uv^\prime\)

Let \(u = x^{2}\), \(u^\prime=2x\); \(v=\sin^{7}x\). To find \(v^\prime\), use chain - rule. Let \(t = \sin x\), \(v=t^{7}\), then \(v^\prime = 7t^{6}\cdot t^\prime=7\sin^{6}x\cos x\).
So \((x^{2}\sin^{7}x)^\prime=2x\sin^{7}x+x^{2}\cdot7\sin^{6}x\cos x=2x\sin^{7}x + 7x^{2}\sin^{6}x\cos x\)

Step2: Differentiate \(x\cos^{- 2}x\) using product rule \((uv)^\prime = u^\prime v+uv^\prime\)

Let \(u = x\), \(u^\prime = 1\); \(v=\cos^{-2}x\). To find \(v^\prime\), use chain - rule. Let \(t=\cos x\), \(v = t^{-2}\), then \(v^\prime=-2t^{-3}\cdot t^\prime=-2\cos^{-3}x(-\sin x)=\frac{2\sin x}{\cos^{3}x}\)
So \((x\cos^{-2}x)^\prime=\cos^{-2}x+x\cdot\frac{2\sin x}{\cos^{3}x}=\frac{1}{\cos^{2}x}+\frac{2x\sin x}{\cos^{3}x}\)

Step3: Find \(\frac{dy}{dx}\)

\(\frac{dy}{dx}=(x^{2}\sin^{7}x)^\prime+(x\cos^{-2}x)^\prime\)
\(\frac{dy}{dx}=2x\sin^{7}x + 7x^{2}\sin^{6}x\cos x+\frac{1}{\cos^{2}x}+\frac{2x\sin x}{\cos^{3}x}\)

Answer:

\(2x\sin^{7}x + 7x^{2}\sin^{6}x\cos x+\frac{1}{\cos^{2}x}+\frac{2x\sin x}{\cos^{3}x}\)