Let y = sin^-1(2x/(1+x^2)) + cos^-1((1-x^2)/(1+x^2)). Find dy/dx and evaluate it at x=1/2. Give the final numerical value.

Let $y = sin^{- 1} (\frac{2 x}{1 + x ^{2}}) + cos^{- 1} (\frac{1 - x ^{2}}{1 + x ^{2}})$ . Find $\frac{d y}{d x}$ and evaluate it at $x = \frac{1}{2}$ . Give the final numerical value.

hard 3 marks Class 12 · Mathematics · Inverse Trigonometric Functions

Answer

The correct answer is 0 — Therefore at $x = \frac{1}{2}$ , $\frac{d y}{d x} = 0$ .

Denote $u = \frac{2 x}{1 + x ^{2}}$ and $v = \frac{1 - x ^{2}}{1 + x ^{2}}$ .
Notice that $u^{2} + v^{2} = (\frac{2 x}{1 + x ^{2}})^{2} + (\frac{1 - x ^{2}}{1 + x ^{2}})^{2} = 1$ , so $v = 1 - u^{2}$ (with $v \geq 0$ for $∣ x ∣ \leq 1$ ).
Recall the identity $sin^{- 1} u + cos^{- 1} v = \frac{π}{2}$ whenever $u^{2} + v^{2} = 1$ and $v \geq 0$ .
Since $∣ x ∣ \leq 1$ includes $x = \frac{1}{2}$ , we have $y = \frac{π}{2}$ , a constant.
Differentiating a constant gives $\frac{d y}{d x} = 0$ for all $x$ in the domain.
Therefore at $x = \frac{1}{2}$ , $\frac{d y}{d x} = 0$ .

Common mistake: Differentiating each term separately and missing the simplification

sin^{- 1} u + cos^{- 1} v = \frac{π}{2}

Concept tested: Derivative of a sum of inverse trig functions using identity