Question

Find the differential coefficient of \( \tan^{-1}\left( \frac{2x}{1-x^2} \right) \) \text{with respect to} \( x \).

Hint:

For inverse trigonometric functions, the chain rule and simplifying the denominator is crucial for differentiation.

Answer 1

Step 1: Let the expression be \( y = \tan^{-1}\left( \frac{2x}{1-x^2} \right) \).
Using the chain rule, we differentiate with respect to \( x \).
\[ \frac{dy}{dx} = \frac{1}{1 + \left( \frac{2x}{1-x^2} \right)^2} . \frac{d}{dx}\left( \frac{2x}{1-x^2} \right) \]
Step 2: Differentiating \( \frac{2x}{1-x^2} \).
Using the quotient rule: \[ \frac{d}{dx}\left( \frac{2x}{1-x^2} \right) = \frac{(1-x^2) . 2 - 2x . (-2x)}{(1-x^2)^2} = \frac{2 - 2x^2 + 4x^2}{(1-x^2)^2} = \frac{2 + 2x^2}{(1-x^2)^2} \]
Step 3: Simplifying the expression.
Substitute this into the derivative: \[ \frac{dy}{dx} = \frac{1}{1 + \left( \frac{2x}{1-x^2} \right)^2} . \frac{2 + 2x^2}{(1-x^2)^2} \] Now, simplify the denominator. We know that: \[ 1 + \left( \frac{2x}{1-x^2} \right)^2 = \frac{(1-x^2)^2 + 4x^2}{(1-x^2)^2} \] Thus, the derivative is: \[ \frac{dy}{dx} = \frac{2 + 2x^2}{(1-x^2)^2 + 4x^2} \]
Final Answer:
\[ \boxed{\frac{dy}{dx} = \frac{2 + 2x^2}{(1-x^2)^2 + 4x^2}} \]