3. [0/6.5 Points] DETAILS MY NOTES SCALCET9 3.XP.5.AE.001. PREVIOUS ANSWERS ASK YOUR TEACHER PRACTICE ANOTHER Example 5 Video Example Differentiate. (a) $$ y=\frac{8}{\sin ^{-1}(x)} $$ (b) $$ f(x)=x \arctan (\sqrt{x}) $$ Solution (a) $$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[8\left(\sin ^{-1}(x)\right)^{-1}\right]=\left(-\frac{8 \sqrt{1-x^{2}}}{\operatorname{arcin}^{2}(x)\left(1-x^{2}\right)}\right. \ & ) \frac{d}{d x}\left(\sin ^{-1}(x)\right) \ & =\square \end{aligned} $$ (b) $$ \begin{aligned} f^{\prime}(x) & =x(\sqrt[{\operatorname{arcan}(\sqrt{x}})]{x})+\arctan (\sqrt{x}) \ & =\frac{\sqrt{x}}{2(x+1)} \end{aligned} $$

Question

UpStudy AI · Accepted Answer

Let's differentiate the given functions step by step.

### (a) Differentiate $$ y = \frac{8}{\sin^{-1}(x)} $$

To differentiate this function, we can use the quotient rule or rewrite it as $$ y = 8 \left( \sin^{-1}(x) \right)^{-1} $$ and then apply the chain rule.

1. Rewrite the function:
   $$
   y = 8 \left( \sin^{-1}(x) \right)^{-1}
   $$

2. Differentiate using the chain rule:
   $$
   \frac{dy}{dx} = 8 \cdot \left(-1\right) \cdot \left( \sin^{-1}(x) \right)^{-2} \cdot \frac{d}{dx} \left( \sin^{-1}(x) \right)
   $$

3. The derivative of $$ \sin^{-1}(x) $$ is:
   $$
   \frac{d}{dx} \left( \sin^{-1}(x) \right) = \frac{1}{\sqrt{1 - x^2}}
   $$

4. Substitute this back into the derivative:
   $$
   \frac{dy}{dx} = -8 \cdot \left( \sin^{-1}(x) \right)^{-2} \cdot \frac{1}{\sqrt{1 - x^2}}
   $$

Thus, the final result for part (a) is:
$$
\frac{dy}{dx} = -\frac{8}{\sin^{-1}(x)^2 \sqrt{1 - x^2}}
$$

### (b) Differentiate $$ f(x) = x \arctan(\sqrt{x}) $$

For this function, we will use the product rule.

1. Identify the two functions:
   - $$ u = x $$
   - $$ v = \arctan(\sqrt{x}) $$

2. Differentiate each part:
   - $$ \frac{du}{dx} = 1 $$
   - To differentiate $$ v $$, we use the chain rule:
     $$
     \frac{dv}{dx} = \frac{1}{1 + (\sqrt{x})^2} \cdot \frac{d}{dx}(\sqrt{x}) = \frac{1}{1 + x} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x}(1 + x)}
     $$

3. Apply the product rule:
   $$
   f'(x) = u \frac{dv}{dx} + v \frac{du}{dx}
   $$
   $$
   f'(x) = x \cdot \frac{1}{2\sqrt{x}(1 + x)} + \arctan(\sqrt{x}) \cdot 1
   $$

4. Simplify the first term:
   $$
   f'(x) = \frac{x}{2\sqrt{x}(1 + x)} + \arctan(\sqrt{x})
   $$
   $$
   = \frac{\sqrt{x}}{2(1 + x)} + \arctan(\sqrt{x})
   $$

Thus, the final result for part (b) is:
$$
f'(x) = \frac{\sqrt{x}}{2(1 + x)} + \arctan(\sqrt{x})
$$
(a) $$ \frac{dy}{dx} = -\frac{8}{\sin^{-1}(x)^2 \sqrt{1 - x^2}} $$ (b) $$ f'(x) = \frac{\sqrt{x}}{2(1 + x)} + \arctan(\sqrt{x}) $$

[0/6.5 Points]
DETAILS
MY NOTES
SCALCET9 3.XP.5.AE.001.
PREVIOUS ANSWERS
ASK YOUR TEACHER
PRACTICE ANOTHER

Example 5
Video Example
Differentiate.
(a)
(b)

Solution
(a)

(b)

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Responder

Solución

(a) Differentiate

(b) Differentiate

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[0/6.5 Points] DETAILS MY NOTES SCALCET9 3.XP.5.AE.001. PREVIOUS ANSWERS ASK YOUR TEACHER PRACTICE ANOTHER Example 5 Video Example Differentiate. (a) (b) Solution (a) (b)