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

Answer

To differentiate the expression \( u = \tan^{-1}\left(\frac{\sqrt{1 + a^2 x^2} - 1}{ax}\right) \) with respect to \( v = \tan^{-1}(ax) \), we will follow these steps:

### Step 1: Define the Variables
Let:
- \( u = \tan^{-1}\left(\frac{\sqrt{1 + a^2 x^2} - 1}{ax}\right) \)
- \( v = \tan^{-1}(ax) \)

### Step 2: Use the Chain Rule
We need to find \( \frac{du}{dv} \). By the chain rule, we have:
\[ \frac{du}{dv} = \frac{du}{dx} \cdot \frac{dx}{dv} \]
This means we need to find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

### Step 3: Differentiate \( u \) with Respect to \( x \)
To differentiate \( u \), we first simplify the expression:
\[ u = \tan^{-1}\left(\frac{\sqrt{1 + a^2 x^2} - 1}{ax}\right) \]
Using the derivative of \( \tan^{-1}(x) \):
\[ \frac{d}{dx} \tan^{-1}(x) = \frac{1}{1 + x^2} \]
Let \( y = \frac{\sqrt{1 + a^2 x^2} - 1}{ax} \). Then:
\[ \frac{du}{dx} = \frac{1}{1 + y^2} \cdot \frac{dy}{dx} \]
Now we need to find \( \frac{dy}{dx} \).

### Step 4: Differentiate \( y \) with Respect to \( x \)
Using the quotient rule:
\[ y = \frac{\sqrt{1 + a^2 x^2} - 1}{ax} \]
Let \( f(x) = \sqrt{1 + a^2 x^2} - 1 \) and \( g(x) = ax \). Then:
\[ \frac{dy}{dx} = \frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2} \]
Where:
- \( f'(x) = \frac{a^2 x}{\sqrt{1 + a^2 x^2}} \)
- \( g'(x) = a \)

Thus,
\[ \frac{dy}{dx} = \frac{ax \cdot \frac{a^2 x}{\sqrt{1 + a^2 x^2}} - \left(\sqrt{1 + a^2 x^2} - 1\right) a}{(ax)^2} \]

### Step 5: Differentiate \( v \) with Respect to \( x \)
Now differentiate \( v \):
\[ v = \tan^{-1}(ax) \]
Using the derivative:
\[ \frac{dv}{dx} = \frac{1}{1 + (ax)^2} \cdot a = \frac{a}{1 + a^2 x^2} \]

### Step 6: Combine the Results
Now we can substitute \( \frac{du}{dx} \) and \( \frac{dv}{dx} \) back into the chain rule:
\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} \]

### Step 7: Final Expression
After substituting and simplifying, we find:
\[ \frac{du}{dv} = \frac{1}{2} \]

### Conclusion
Thus, the derivative of \( u \) with respect to \( v \) is:
\[ \frac{du}{dv} = \frac{1}{2} \]