Question

use the product rule to find the derivative of the function. h(t) = √t(6 - t²)

Explanation:

Step1: Identify functions u and v

Let $u = \sqrt{t}=t^{\frac{1}{2}}$ and $v = 6 - t^{2}$.

Step2: Find derivatives of u and v

The derivative of $u=t^{\frac{1}{2}}$ is $u'=\frac{1}{2}t^{-\frac{1}{2}}$ using the power - rule $\frac{d}{dt}(t^n)=nt^{n - 1}$. The derivative of $v = 6 - t^{2}$ is $v'=-2t$.

Step3: Apply the product rule

The product rule states that $(uv)'=u'v + uv'$. Substitute $u$, $u'$, $v$, and $v'$ into the product rule:
\[

$$\begin{align*} h'(t)&=u'v+uv'\\ &=\frac{1}{2}t^{-\frac{1}{2}}(6 - t^{2})+t^{\frac{1}{2}}(-2t)\\ &=\frac{6}{2}t^{-\frac{1}{2}}-\frac{1}{2}t^{-\frac{1}{2}}t^{2}-2t^{\frac{3}{2}}\\ &=3t^{-\frac{1}{2}}-\frac{1}{2}t^{\frac{3}{2}}-2t^{\frac{3}{2}}\\ &=3t^{-\frac{1}{2}}-\frac{5}{2}t^{\frac{3}{2}} \end{align*}$$

\]

Answer:

$h'(t)=3t^{-\frac{1}{2}}-\frac{5}{2}t^{\frac{3}{2}}$