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(6 - 2t) \\)\\( \times \\)\
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Explanation:

Step1: Rewrite function with exponents

Rewrite $\sqrt{t}$ as $t^{\frac{1}{2}}$:
$$h(t) = t^{\frac{1}{2}}(6 - t^2)$$

Step2: Define product rule terms

Let $f(t) = t^{\frac{1}{2}}$, $g(t) = 6 - t^2$. Product rule: $h'(t) = f'(t)g(t) + f(t)g'(t)$

Step3: Compute $f'(t)$ and $g'(t)$

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

Step4: Substitute into product rule

$$\begin{align*} h'(t) &= \frac{1}{2}t^{-\frac{1}{2}}(6 - t^2) + t^{\frac{1}{2}}(-2t)\\ &= \frac{6 - t^2}{2t^{\frac{1}{2}}} - 2t^{\frac{3}{2}} \end{align*}$$

Step5: Simplify the expression

Combine terms over common denominator $2t^{\frac{1}{2}}$:

$$\begin{align*} h'(t) &= \frac{6 - t^2 - 4t^2}{2t^{\frac{1}{2}}}\\ &= \frac{6 - 5t^2}{2\sqrt{t}} \end{align*}$$

Answer:

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