Question

use the product rule to find the derivative of the function.\\( h(t) = sqrt{t}(6 - t^2) \\)\\( h(t) = \frac{1}{2}t^{-left( \frac{1}{2}
ight)}(6 - 2t) \\)\\( \boldsymbol{\times} \\)\
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Explanation:

Step1: Rewrite function with exponents

$h(t) = t^{\frac{1}{2}}(6 - t^2)$

Step2: Recall Product Rule

If $h(t)=f(t)g(t)$, then $h'(t)=f'(t)g(t)+f(t)g'(t)$

Step3: Define $f(t),g(t)$ and find derivatives

Let $f(t)=t^{\frac{1}{2}}$, so $f'(t)=\frac{1}{2}t^{-\frac{1}{2}}$
Let $g(t)=6-t^2$, so $g'(t)=-2t$

Step4: Apply Product Rule

$h'(t)=\frac{1}{2}t^{-\frac{1}{2}}(6 - t^2) + t^{\frac{1}{2}}(-2t)$

Step5: Simplify each term

First term: $\frac{6 - t^2}{2t^{\frac{1}{2}}}$; Second term: $-2t^{\frac{3}{2}}$
Combine terms over common denominator $2t^{\frac{1}{2}}$:
$h'(t)=\frac{6 - t^2 - 4t^2}{2t^{\frac{1}{2}}}=\frac{6 - 5t^2}{2\sqrt{t}}$

Answer:

$h'(t)=\frac{6 - 5t^2}{2\sqrt{t}}$ (or equivalent form $\frac{1}{2}t^{-\frac{1}{2}}(6 - t^2) - 2t^{\frac{3}{2}}$)