The derivative of `tan^(-1)x` w.r.t`cot^(-1)` x is

Options

A. `(pi)/(2)`
B. `-1`
C. 1
D. none of these

Answer

To find the derivative of \( \tan^{-1} x \) with respect to \( \cot^{-1} x \), we can follow these steps:

### Step 1: Set up the relationship
Let:
- \( y = \tan^{-1} x \)
- \( z = \cot^{-1} x \)

We want to find \( \frac{dy}{dz} \).

### Step 2: Use the chain rule
According to the chain rule, we can express the derivative as:
\[ \frac{dy}{dz} = \frac{dy}{dx} \cdot \frac{dx}{dz} \]
This means we need to find \( \frac{dy}{dx} \) and \( \frac{dz}{dx} \).

### Step 3: Differentiate \( y = \tan^{-1} x \)
The derivative of \( y \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{1}{1 + x^2} \]

### Step 4: Differentiate \( z = \cot^{-1} x \)
The derivative of \( z \) with respect to \( x \) is:
\[ \frac{dz}{dx} = -\frac{1}{1 + x^2} \]

### Step 5: Substitute into the chain rule formula
Now we can substitute \( \frac{dy}{dx} \) and \( \frac{dz}{dx} \) back into our equation:
\[ \frac{dy}{dz} = \frac{dy}{dx} \cdot \frac{dx}{dz} = \frac{dy}{dx} \cdot \frac{1}{\frac{dz}{dx}} = \frac{\frac{1}{1 + x^2}}{-\frac{1}{1 + x^2}} \]

### Step 6: Simplify the expression
This simplifies to:
\[ \frac{dy}{dz} = \frac{1}{1 + x^2} \cdot \left(-\frac{1 + x^2}{1}\right) = -1 \]

### Final Answer
Thus, the derivative of \( \tan^{-1} x \) with respect to \( \cot^{-1} x \) is:
\[ \frac{d(\tan^{-1} x)}{d(\cot^{-1} x)} = -1 \]

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