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

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

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

Answer

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: [ \begin{align*} \frac{dy}{dx}&=12x\sin x+6x^{2}\cos x + 12\cos x-12x\sin x-12\cos x\ &=6x^{2}\cos x \end{align*} ]

Answer:

(6x^{2}\cos x)