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

Answer

To differentiate \( \cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) \) with respect to \( \tan^{-1}(x) \), we will follow these steps:

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

### Step 2: Differentiate \( u \) with respect to \( x \)
To find \( \frac{du}{dx} \), we need to differentiate \( u \):
Using the chain rule:
\[ \frac{du}{dx} = -\frac{1}{\sqrt{1 - \left(\frac{1}{\sqrt{1+x^2}}\right)^2}} \cdot \frac{d}{dx}\left(\frac{1}{\sqrt{1+x^2}}\right) \]

First, simplify \( \sqrt{1 - \left(\frac{1}{\sqrt{1+x^2}}\right)^2} \):
\[ \sqrt{1 - \frac{1}{1+x^2}} = \sqrt{\frac{(1+x^2)-1}{1+x^2}} = \sqrt{\frac{x^2}{1+x^2}} = \frac{|x|}{\sqrt{1+x^2}} \]

Now, differentiate \( \frac{1}{\sqrt{1+x^2}} \):
\[ \frac{d}{dx}\left(\frac{1}{\sqrt{1+x^2}}\right) = -\frac{1}{2(1+x^2)^{3/2}} \cdot 2x = -\frac{x}{(1+x^2)^{3/2}} \]

Thus,
\[ \frac{du}{dx} = -\frac{1}{\frac{|x|}{\sqrt{1+x^2}}} \cdot \left(-\frac{x}{(1+x^2)^{3/2}}\right) = \frac{x}{|x|\sqrt{1+x^2}} \cdot \frac{1}{(1+x^2)^{3/2}} = \frac{x}{|x|(1+x^2)} = \frac{1}{(1+x^2)} \quad \text{(for } x > 0\text{)} \]

### Step 3: Differentiate \( v \) with respect to \( x \)
Now, differentiate \( v \):
\[ \frac{dv}{dx} = \frac{1}{1+x^2} \]

### Step 4: Use the chain rule to find \( \frac{du}{dv} \)
Using the chain rule:
\[ \frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{\frac{1}{1+x^2}}{\frac{1}{1+x^2}} = 1 \]

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

### Final Answer
The final answer is:
\[ \frac{d}{d(\tan^{-1}(x))} \cos^{-1}\left(\frac{1}{\sqrt{1+x^2}}\right) = 1 \]

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