Question

question 19
find \\( \frac{d}{dx} \left x^3 \sin x \
ight \\).
\\( \bigcirc \\ x^3 \cos x - 3x^2 \sin x \\)
\\( \bigcirc \\ 3x^2 \sin x - x^3 \cos x \\)
\\( \bigcirc \\ 3x^2 \sin x + x^3 \cos x \\)
\\( \bigcirc \\ 3x^2 \cos x - x^3 \sin x \\)
\\( \bigcirc \\) no correct answer choice is given.

Explanation:

Step1: Recall Product Rule

The product rule for differentiation is \(\frac{d}{dx}[u(x)v(x)] = u'(x)v(x)+u(x)v'(x)\). Let \(u(x)=x^{3}\) and \(v(x)=\sin x\).

Step2: Differentiate \(u(x)\) and \(v(x)\)

Differentiate \(u(x) = x^{3}\) using the power rule \(\frac{d}{dx}[x^{n}]=nx^{n - 1}\), so \(u'(x)=3x^{2}\).
Differentiate \(v(x)=\sin x\), we know that \(\frac{d}{dx}[\sin x]=\cos x\), so \(v'(x)=\cos x\).

Step3: Apply Product Rule

Substitute \(u(x)\), \(u'(x)\), \(v(x)\) and \(v'(x)\) into the product rule:
\(\frac{d}{dx}[x^{3}\sin x]=u'(x)v(x)+u(x)v'(x)=3x^{2}\sin x + x^{3}\cos x\)

Answer:

\(3x^{2}\sin x + x^{3}\cos x\) (the third option: \(3x^{2}\sin x + x^{3}\cos x\))