Question

find $\frac{dy}{dx}$ for $y = 6x^{2}\sin x + 12x\cos x - 12\sin x$.
$\frac{dy}{dx}=$

Explanation:

Step1: Differentiate each term

Use the product rule \((uv)^\prime = u^\prime v+uv^\prime\) and \((\sin x)^\prime=\cos x\), \((\cos x)^\prime =-\sin x\), \((x^n)^\prime=nx^{n - 1}\).
For \(y_1 = 6x^{2}\sin x\), let \(u = 6x^{2}\), \(v=\sin x\). Then \(u^\prime=12x\), \(v^\prime=\cos x\). So \((6x^{2}\sin x)^\prime=12x\sin x+6x^{2}\cos x\).
For \(y_2 = 12x\cos x\), let \(u = 12x\), \(v=\cos x\). Then \(u^\prime = 12\), \(v^\prime=-\sin x\). So \((12x\cos x)^\prime=12\cos x-12x\sin x\).
For \(y_3=-12\sin x\), \((- 12\sin x)^\prime=-12\cos x\).

Step2: Combine the derivatives

\(\frac{dy}{dx}=(12x\sin x + 6x^{2}\cos x)+(12\cos x-12x\sin x)-12\cos x\)
Simplify the expression:

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Answer:

\(6x^{2}\cos x\)