Question

find $\frac{dy}{dx}$ for $y = 7x^{2}sin x+14xcos x - 14sin x$. $\frac{dy}{dx}=square$

Explanation:

Step1: Apply sum - difference rule of derivatives

$\frac{dy}{dx}=\frac{d}{dx}(7x^{2}\sin x)+\frac{d}{dx}(14x\cos x)-\frac{d}{dx}(14\sin x)$

Step2: Use product rule for $\frac{d}{dx}(7x^{2}\sin x)$

Product rule: $(uv)^\prime = u^\prime v+uv^\prime$. Let $u = 7x^{2}$ and $v=\sin x$. Then $\frac{d}{dx}(7x^{2}\sin x)=14x\sin x + 7x^{2}\cos x$

Step3: Use product rule for $\frac{d}{dx}(14x\cos x)$

Let $u = 14x$ and $v=\cos x$. Then $\frac{d}{dx}(14x\cos x)=14\cos x-14x\sin x$

Step4: Differentiate $\frac{d}{dx}(14\sin x)$

$\frac{d}{dx}(14\sin x)=14\cos x$

Step5: Combine the results

$\frac{dy}{dx}=(14x\sin x + 7x^{2}\cos x)+(14\cos x-14x\sin x)-14\cos x$
$=7x^{2}\cos x$

Answer:

$7x^{2}\cos x$