Question

1. 0 / 5.55 points

differentiate.
$g(t) = t^6 \cos t$
$g(t)$

1. - / 5.55 points

differentiate.
$f(x) = 3\sqrt{x} \sin x$
$f(x) = $

Explanation:

Step1: Apply product rule to \(g(t)\)

Product rule: \((uv)' = u'v + uv'\), where \(u=t^6\), \(v=\cos t\)

Step2: Compute derivatives of \(u\) and \(v\)

\(u' = 6t^5\), \(v' = -\sin t\)

Step3: Substitute into product rule

\(g'(t) = 6t^5\cos t + t^6(-\sin t)\)

Step4: Rewrite \(f(x)\) for differentiation

\(f(x)=3x^{1/2}\sin x\), apply product rule (\(u=3x^{1/2}\), \(v=\sin x\))

Step5: Compute derivatives of \(u\) and \(v\)

\(u' = 3*(1/2)x^{-1/2} = \frac{3}{2\sqrt{x}}\), \(v' = \cos x\)

Step6: Substitute into product rule

\(f'(x) = \frac{3}{2\sqrt{x}}\sin x + 3\sqrt{x}\cos x\)

Answer:

\(g'(t) = 6t^5\cos t - t^6\sin t\)
\(f'(x) = \frac{3\sin x}{2\sqrt{x}} + 3\sqrt{x}\cos x\)