Question

find \\( \frac { d y } { d x } \\)
\\( y = x ^ { 2 } \cos x \\)
\\( \frac { d y } { d x } = \\)

Explanation:

Step1: Apply the product rule

The product rule states that if \(y = u\cdot v\), then \(\frac{dy}{dx}=u'\cdot v + u\cdot v'\). Let \(u = x^{2}\) and \(v=\cos x\). First, find \(u'\) and \(v'\).
For \(u = x^{2}\), using the power rule \(\frac{d}{dx}(x^{n})=nx^{n - 1}\), we have \(u'=\frac{d}{dx}(x^{2}) = 2x\).
For \(v=\cos x\), using the derivative formula \(\frac{d}{dx}(\cos x)=-\sin x\), we have \(v'=-\sin x\).

Step2: Substitute into the product rule formula

Substitute \(u = x^{2}\), \(u' = 2x\), \(v=\cos x\), and \(v'=-\sin x\) into \(\frac{dy}{dx}=u'\cdot v+u\cdot v'\).
We get \(\frac{dy}{dx}=2x\cdot\cos x+x^{2}\cdot(-\sin x)\).

Step3: Simplify the expression

\(\frac{dy}{dx}=2x\cos x - x^{2}\sin x\).

Answer:

\(2x\cos x - x^{2}\sin x\)