If `y= tan^-1((1-x)/(1+x))+cot^-1((1-x)/(1+x))` then `dy/dx`= ?

Options

A. -1
B. 1
C. 0
D. None of these

Answer

To find \( \frac{dy}{dx} \) for the function

\[ y = \tan^{-1}\left(\frac{1-x}{1+x}\right) + \cot^{-1}\left(\frac{1-x}{1+x}\right), \]

we can use the property of inverse trigonometric functions.

### Step 1: Simplify the expression

We know that:

\[ \tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2} \]

for any \( x \). In our case, let \( x = \frac{1-x}{1+x} \). Thus, we can rewrite \( y \):

\[ y = \frac{\pi}{2} \]

### Step 2: Differentiate the constant

Now, we differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{2}\right) \]

Since \( \frac{\pi}{2} \) is a constant, its derivative is:

\[ \frac{dy}{dx} = 0 \]

### Final Answer

Thus, the derivative \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = 0 \]

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