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Exercise 5.3 · Q9

Q.Find dydx\frac{dy}{dx} in the following: y=sin⁡−1(2x1+x2)y = \sin^{-1} \left(\frac{2x}{1+x^2}\right)

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Substituting x=tan⁡θx = \tan\theta reduces yy to a piecewise expression in tan⁡−1x\tan^{-1}x; differentiating each branch gives dydx=21+x2\dfrac{dy}{dx} = \dfrac{2}{1+x^2} for ∣x∣<1|x|<1 and dydx=−21+x2\dfrac{dy}{dx} = -\dfrac{2}{1+x^2} for ∣x∣>1|x|>1.

Why the Substitution Helps

The argument 2x1+x2\dfrac{2x}{1+x^2} is exactly the double-angle formula sin⁡2θ=2tan⁡θ1+tan⁡2θ\sin 2\theta = \dfrac{2\tan\theta}{1+\tan^2\theta} in disguise. Setting x=tan⁡θx = \tan\theta turns the messy rational expression inside sin⁡−1\sin^{-1} into a clean sin⁡2θ\sin 2\theta, but simplifying sin⁡−1(sin⁡2θ)\sin^{-1}(\sin 2\theta) to 2θ2\theta is only valid when 2θ2\theta lies in the principal range of sin⁡−1\sin^{-1} — otherwise a correction of ±π\pm\pi is required. Getting this piecewise form right before differentiating is essential, since the derivative differs by sign across the pieces.

Step-by-Step Solution

1. Substitute x=tan⁡θx = \tan\theta.

Let θ=tan⁡−1x\theta = \tan^{-1}x, so θ∈(−π2,π2)\theta \in \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right). Then:

2x1+x2=2tan⁡θ1+tan⁡2θ=sin⁡2θ⇒y=sin⁡−1(sin⁡2θ).\frac{2x}{1+x^2} = \frac{2\tan\theta}{1+\tan^2\theta} = \sin 2\theta \quad\Rightarrow\quad y = \sin^{-1}(\sin 2\theta).

2. Determine when sin⁡−1(sin⁡2θ)=2θ\sin^{-1}(\sin 2\theta) = 2\theta directly.

This holds only when 2θ∈[−π2,π2]2\theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right], i.e. θ∈[−π4,π4]\theta \in \left[-\tfrac{\pi}{4}, \tfrac{\pi}{4}\right], i.e. x=tan⁡θ∈[−1,1]x = \tan\theta \in [-1, 1].

  • Case ∣x∣≤1|x| \le 1: 2θ∈[−π2,π2]2\theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right], so y=2θ=2tan⁡−1xy = 2\theta = 2\tan^{-1}x.
  • Case x>1x > 1: θ∈(π4,π2)\theta \in \left(\tfrac{\pi}{4}, \tfrac{\pi}{2}\right), so 2θ∈(π2,π)2\theta \in \left(\tfrac{\pi}{2}, \pi\right). Using sin⁡(π−2θ)=sin⁡2θ\sin(\pi - 2\theta) = \sin 2\theta and π−2θ∈(0,π2)\pi - 2\theta \in \left(0, \tfrac{\pi}{2}\right) (inside the principal range), y=π−2θ=π−2tan⁡−1xy = \pi - 2\theta = \pi - 2\tan^{-1}x.
  • Case x<−1x < -1: θ∈(−π2,−π4)\theta \in \left(-\tfrac{\pi}{2}, -\tfrac{\pi}{4}\right), so 2θ∈(−π,−π2)2\theta \in \left(-\pi, -\tfrac{\pi}{2}\right). Using sin⁡(−π−2θ)=sin⁡2θ\sin(-\pi - 2\theta) = \sin 2\theta and −π−2θ∈(−π2,0)-\pi - 2\theta \in \left(-\tfrac{\pi}{2}, 0\right) (inside the principal range), y=−π−2θ=−π−2tan⁡−1xy = -\pi - 2\theta = -\pi - 2\tan^{-1}x.

So:

y={2tan⁡−1x,∣x∣≤1π−2tan⁡−1x,x>1−π−2tan⁡−1x,x<−1y = \begin{cases} 2\tan^{-1}x, & |x| \le 1 \\ \pi - 2\tan^{-1}x, & x > 1 \\ -\pi - 2\tan^{-1}x, & x < -1 \end{cases}

3. Differentiate each branch. …

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