Differentiate w.r.t. as indicated :
`sqrt(1+x^(2))" w.r.t. "tan^(-1)x`

Answer

To differentiate \( \sqrt{1+x^2} \) with respect to \( \tan^{-1}x \), we will use the formula for the derivative of one function with respect to another, which is given by:

\[ \frac{dy}{dx} = \frac{f'(x)}{g'(x)} \]

where \( f(x) = \sqrt{1+x^2} \) and \( g(x) = \tan^{-1}x \).

### Step 1: Differentiate \( f(x) = \sqrt{1+x^2} \)

To find \( f'(x) \):

\[ f(x) = (1+x^2)^{1/2} \]

Using the chain rule:

\[ f'(x) = \frac{1}{2}(1+x^2)^{-1/2} \cdot (2x) = \frac{x}{\sqrt{1+x^2}} \]

### Step 2: Differentiate \( g(x) = \tan^{-1}x \)

To find \( g'(x) \):

\[ g'(x) = \frac{1}{1+x^2} \]

### Step 3: Find \( \frac{f'(x)}{g'(x)} \)

Now we can find the derivative of \( \sqrt{1+x^2} \) with respect to \( \tan^{-1}x \):

\[ \frac{dy}{dx} = \frac{f'(x)}{g'(x)} = \frac{\frac{x}{\sqrt{1+x^2}}}{\frac{1}{1+x^2}} \]

This simplifies to:

\[ \frac{dy}{dx} = \frac{x}{\sqrt{1+x^2}} \cdot (1+x^2) = \frac{x(1+x^2)}{\sqrt{1+x^2}} \]

### Step 4: Simplify the expression

The expression simplifies to:

\[ \frac{dy}{dx} = x \sqrt{1+x^2} \]

### Final Answer

Thus, the derivative of \( \sqrt{1+x^2} \) with respect to \( \tan^{-1}x \) is:

\[ \frac{dy}{dx} = x \sqrt{1+x^2} \]

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