Differentiate each of the following with respect to x: a) ^-1 5x, b) ^-1 3x, c) ^-1 sqrt(2x)

Here are the solutions for the differentiation and integration problems.

1. Differentiate each of the following with respect to $x$.

a) $\sin^{-1} 5x$

Step 1: Use the chain rule with $\frac{d}{dx} \sin^{-1}(u) = \frac{1}{\sqrt{1-u^2}} \frac{du}{dx}$. Here $u=5x$, so $\frac{du}{dx}=5$.

Step 2: Substitute $u$ and $\frac{du}{dx}$ into the formula.

$$\frac{d}{dx} (\sin^{-1} 5x) = \frac{1}{\sqrt{1-(5x)^2}} \cdot 5$$

Step 3: Simplify the expression.

$$= \frac{5}{\sqrt{1-25x^2}}$$

The derivative is $\boxed{\frac{5}{\sqrt{1-25x^2}}}$.

b) $\tan^{-1} 3x$

Step 1: Use the chain rule with $\frac{d}{dx} \tan^{-1}(u) = \frac{1}{1+u^2} \frac{du}{dx}$. Here $u=3x$, so $\frac{du}{dx}=3$.

Step 2: Substitute $u$ and $\frac{du}{dx}$ into the formula.

$$\frac{d}{dx} (\tan^{-1} 3x) = \frac{1}{1+(3x)^2} \cdot 3$$

Step 3: Simplify the expression.

$$= \frac{3}{1+9x^2}$$

The derivative is $\boxed{\frac{3}{1+9x^2}}$.

c) $\sin^{-1} \sqrt{2x}$

Step 1: Use the chain rule with $\frac{d}{dx} \sin^{-1}(u) = \frac{1}{\sqrt{1-u^2}} \frac{du}{dx}$. Here $u=\sqrt{2x}=(2x)^{1/2}$.

Step 2: Find $\frac{du}{dx}$ using the chain rule: $\frac{du}{dx} = \frac{1}{2}(2x)^{-1/2} \cdot 2 = (2x)^{-1/2} = \frac{1}{\sqrt{2x}}$.

Step 3: Substitute $u$ and $\frac{du}{dx}$ into the $\sin^{-1}$ formula.

$$\frac{d}{dx} (\sin^{-1} \sqrt{2x}) = \frac{1}{\sqrt{1-(\sqrt{2x})^2}} \cdot \frac{1}{\sqrt{2x}}$$

Step 4: Simplify the expression.

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