Question

If \( y = \tan^2(\cos^{-1}\sqrt{\frac{1+x^2}{2}}) \), then \( \frac{dy}{dx} = \)

• A. \( \frac{4x}{(1-x^2)^2} \)
• B. \( \frac{4x}{(1+x^2)^2} \)
• C. \( \frac{-4x}{(1+x^2)^2} \)
• D. \( \frac{4x}{1+x^2} \)

Hint:

When differentiating complex trigonometric or inverse trigonometric functions, always try to simplify the expression algebraically first. Using trigonometric identities can often transform a complicated function into a much simpler algebraic one, making differentiation significantly easier.

Answer 1

The Correct Option is C

Let's simplify the expression for y first.
Let \( \theta = \cos^{-1}\sqrt{\frac{1+x^2}{2}} \). This means \( \cos\theta = \sqrt{\frac{1+x^2}{2}} \).
Then \( \cos^2\theta = \frac{1+x^2}{2} \).
We want to find \( y = \tan^2\theta \).
Using the identity \( \tan^2\theta = \sec^2\theta - 1 = \frac{1}{\cos^2\theta} - 1 \).
Substitute the expression for \( \cos^2\theta \).
\( y = \frac{1}{(1+x^2)/2} - 1 = \frac{2}{1+x^2} - 1 \).
\( y = \frac{2 - (1+x^2)}{1+x^2} = \frac{1-x^2}{1+x^2} \).
Now, we differentiate this simplified expression for y with respect to x using the quotient rule.
Let \( u = 1-x^2 \) and \( v = 1+x^2 \). Then \( u' = -2x \) and \( v' = 2x \).
\( \frac{dy}{dx} = \frac{u'v - uv'}{v^2} = \frac{(-2x)(1+x^2) - (1-x^2)(2x)}{(1+x^2)^2} \).
\( \frac{dy}{dx} = \frac{-2x - 2x^3 - (2x - 2x^3)}{(1+x^2)^2} \).
\( \frac{dy}{dx} = \frac{-2x - 2x^3 - 2x + 2x^3}{(1+x^2)^2} = \frac{-4x}{(1+x^2)^2} \).