26. [-/5 Points] DETAlLS MY NOTES SCALCET9 3.4.063. A graphing calculator is recommended. Let $$ f(x)=x \sqrt{6-x^{2}} $$. (a) Find $$ f(x) $$. $$ f^{\prime}(x)=\square $$ (b) Check to see that your answer to part (a) is reasonable by comparing the graphs of $$ f $$ and $$ f $$ ".

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We begin with 
$$
f(x)=x \sqrt{6-x^2}
$$
and wish to compute its derivative $$ f'(x) $$.

1. Write $$\sqrt{6-x^2}$$ as $$(6-x^2)^{\frac{1}{2}}$$ so that
   $$
   f(x)=x (6-x^2)^{\frac{1}{2}}.
   $$

2. Use the product rule. If
   $$
   f(x)=u(x)v(x) \quad \text{with} \quad u(x)=x \quad \text{and} \quad v(x)=(6-x^2)^{\frac{1}{2}},
   $$
   then
   $$
   f'(x)=u'(x)v(x)+u(x)v'(x).
   $$

3. Compute $$ u'(x) $$:
   $$
   u'(x)=1.
   $$

4. To differentiate $$v(x)=(6-x^2)^{\frac{1}{2}}$$, use the chain rule. Let
   $$
   g(x)=6-x^2 \quad \text{so that} \quad v(x)=g(x)^{\frac{1}{2}}.
   $$
   Then
   $$
   v'(x)=\frac{1}{2}g(x)^{-\frac{1}{2}}g'(x).
   $$
   Since
   $$
   g'(x)=-2x,
   $$
   we obtain
   $$
   v'(x)=\frac{1}{2}(6-x^2)^{-\frac{1}{2}}(-2x)=-\frac{x}{\sqrt{6-x^2}}.
   $$

5. Now substitute back into the product rule:
   $$
   f'(x)=1\cdot \sqrt{6-x^2}+x\left(-\frac{x}{\sqrt{6-x^2}}\right).
   $$
   Simplify:
   $$
   f'(x)=\sqrt{6-x^2}-\frac{x^2}{\sqrt{6-x^2}}.
   $$

6. To combine the terms, write them with a common denominator:
   $$
   f'(x)=\frac{(6-x^2)-x^2}{\sqrt{6-x^2}}=\frac{6-2x^2}{\sqrt{6-x^2}}.
   $$

Thus, the derivative is:
$$
f'(x)=\frac{6-2x^2}{\sqrt{6-x^2}}.
$$

For part (b), one can use a graphing calculator to graph both $$ f(x) $$ and $$ f'(x) $$. The graph of $$ f(x) $$ will show the original function, while the graph of $$ f'(x) $$ represents the slope of $$ f(x) $$ at any point $$ x $$. By comparing:

- At points where the graph $$ f(x) $$ has a horizontal tangent (i.e. local maximum or minimum), the value of $$ f'(x) $$ should be $$0$$. Notice that setting $$ 6-2x^2=0 $$ leads to $$ x^2=3 $$ (provided $$ \sqrt{6-x^2} \neq 0 $$), which corresponds well with the features of the graph of $$ f(x) $$.
- The sign of $$ f'(x) $$ indicates whether $$ f(x) $$ is increasing or decreasing, and this should match the behavior seen in the graph of $$ f(x) $$.

This graphical verification helps confirm that the derived $$ f'(x) $$ is reasonable.
$$ f'(x) = \frac{6 - 2x^2}{\sqrt{6 - x^2}} $$. For part (b), graphing both $$ f(x) $$ and $$ f'(x) $$ shows that at points where $$ f(x) $$ has horizontal tangents, $$ f'(x) $$ equals zero, and the signs of $$ f'(x) $$ correspond to the increasing or decreasing behavior of $$ f(x) $$.

[-/5 Points] DETAlLS MY NOTES SCALCET9 3.4.063.
A graphing calculator is recommended.
Let .
(a) Find .

(b) Check to see that your answer to part (a) is reasonable by comparing the graphs of and ".

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[-/5 Points] DETAlLS MY NOTES SCALCET9 3.4.063. A graphing calculator is recommended. Let . (a) Find . (b) Check to see that your answer to part (a) is reasonable by comparing the graphs of and ".