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Question: Let $y = 2 \tan^{-1}x$ and $z = \sin^{-1} \frac{2x}{1 + x^2}$, then $\left|\frac{dy}{dz}\right|$ equ...

Let y=2tan⁡−1xy = 2 \tan^{-1}x and z=sin⁡−12x1+x2z = \sin^{-1} \frac{2x}{1 + x^2}, then ∣dydz∣\left|\frac{dy}{dz}\right| equals

A

-1

B

0

C

1

D

2

Answer

1

Explanation

Solution

Let y=2tan⁡−1xy = 2 \tan^{-1}x and z=sin⁡−12x1+x2z = \sin^{-1} \frac{2x}{1 + x^2}. We want to find ∣dydz∣\left|\frac{dy}{dz}\right|.

We can find dydz\frac{dy}{dz} using the chain rule: dydz=dy/dxdz/dx\frac{dy}{dz} = \frac{dy/dx}{dz/dx}.

First, find dydx\frac{dy}{dx}: y=2tan⁡−1xy = 2 \tan^{-1}x dydx=ddx(2tan⁡−1x)=2⋅11+x2=21+x2\frac{dy}{dx} = \frac{d}{dx}(2 \tan^{-1}x) = 2 \cdot \frac{1}{1 + x^2} = \frac{2}{1 + x^2}.

Next, find dzdx\frac{dz}{dx}: z=sin⁡−12x1+x2z = \sin^{-1} \frac{2x}{1 + x^2}. Let x=tan⁡θx = \tan \theta, where θ=tan⁡−1x\theta = \tan^{-1}x. Then 2x1+x2=2tan⁡θ1+tan⁡2θ=sin⁡(2θ)\frac{2x}{1 + x^2} = \frac{2 \tan \theta}{1 + \tan^2 \theta} = \sin(2\theta). So, z=sin⁡−1(sin⁡(2θ))=sin⁡−1(sin⁡(2tan⁡−1x))z = \sin^{-1}(\sin(2\theta)) = \sin^{-1}(\sin(2 \tan^{-1}x)).

The value of sin⁡−1(sin⁡u)\sin^{-1}(\sin u) depends on the range of uu. sin⁡−1(sin⁡u)=u\sin^{-1}(\sin u) = u if u∈[−π2,π2]u \in [-\frac{\pi}{2}, \frac{\pi}{2}]. sin⁡−1(sin⁡u)=π−u\sin^{-1}(\sin u) = \pi - u if u∈[π2,3π2]u \in [\frac{\pi}{2}, \frac{3\pi}{2}]. sin⁡−1(sin⁡u)=−π−u\sin^{-1}(\sin u) = -\pi - u if u∈[−3π2,−π2]u \in [-\frac{3\pi}{2}, -\frac{\pi}{2}]. In our case, u=2tan⁡−1xu = 2 \tan^{-1}x. The range of tan⁡−1x\tan^{-1}x is (−π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}), so the range of 2tan⁡−1x2 \tan^{-1}x is (−π,π)(-\pi, \pi).

We consider different cases for the value of xx:

Case 1: −1<x<1-1 < x < 1. If −1<x<1-1 < x < 1, then −π4<tan⁡−1x<π4-\frac{\pi}{4} < \tan^{-1}x < \frac{\pi}{4}. Then −π2<2tan⁡−1x<π2-\frac{\pi}{2} < 2 \tan^{-1}x < \frac{\pi}{2}. In this range, z=sin⁡−1(sin⁡(2tan⁡−1x))=2tan⁡−1xz = \sin^{-1}(\sin(2 \tan^{-1}x)) = 2 \tan^{-1}x. So, z=yz = y. dzdx=ddx(2tan⁡−1x)=21+x2\frac{dz}{dx} = \frac{d}{dx}(2 \tan^{-1}x) = \frac{2}{1 + x^2}. Then dydz=dy/dxdz/dx=2/(1+x2)2/(1+x2)=1\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{2/(1+x^2)}{2/(1+x^2)} = 1.

Case 2: x>1x > 1. If x>1x > 1, then π4<tan⁡−1x<π2\frac{\pi}{4} < \tan^{-1}x < \frac{\pi}{2}. Then π2<2tan⁡−1x<π\frac{\pi}{2} < 2 \tan^{-1}x < \pi. In this range, u=2tan⁡−1x∈(π2,π)u = 2 \tan^{-1}x \in (\frac{\pi}{2}, \pi). This range is part of [π2,3π2][\frac{\pi}{2}, \frac{3\pi}{2}]. z=sin⁡−1(sin⁡(2tan⁡−1x))=π−2tan⁡−1xz = \sin^{-1}(\sin(2 \tan^{-1}x)) = \pi - 2 \tan^{-1}x. So, z=π−yz = \pi - y. dzdx=ddx(π−2tan⁡−1x)=0−21+x2=−21+x2\frac{dz}{dx} = \frac{d}{dx}(\pi - 2 \tan^{-1}x) = 0 - \frac{2}{1 + x^2} = -\frac{2}{1 + x^2}. Then dydz=dy/dxdz/dx=2/(1+x2)−2/(1+x2)=−1\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{2/(1+x^2)}{-2/(1+x^2)} = -1.

Case 3: x<−1x < -1. If x<−1x < -1, then −π2<tan⁡−1x<−π4-\frac{\pi}{2} < \tan^{-1}x < -\frac{\pi}{4}. Then −π<2tan⁡−1x<−π2-\pi < 2 \tan^{-1}x < -\frac{\pi}{2}. In this range, u=2tan⁡−1x∈(−π,−π2)u = 2 \tan^{-1}x \in (-\pi, -\frac{\pi}{2}). This range is part of [−3π2,−π2][-\frac{3\pi}{2}, -\frac{\pi}{2}]. z=sin⁡−1(sin⁡(2tan⁡−1x))=−π−2tan⁡−1xz = \sin^{-1}(\sin(2 \tan^{-1}x)) = -\pi - 2 \tan^{-1}x. So, z=−π−yz = -\pi - y. dzdx=ddx(−π−2tan⁡−1x)=0−21+x2=−21+x2\frac{dz}{dx} = \frac{d}{dx}(-\pi - 2 \tan^{-1}x) = 0 - \frac{2}{1 + x^2} = -\frac{2}{1 + x^2}. Then dydz=dy/dxdz/dx=2/(1+x2)−2/(1+x2)=−1\frac{dy}{dz} = \frac{dy/dx}{dz/dx} = \frac{2/(1+x^2)}{-2/(1+x^2)} = -1.

The derivative dzdx\frac{dz}{dx} is not defined at x=1x=1 and x=−1x=-1 because the function z=sin⁡−12x1+x2z = \sin^{-1} \frac{2x}{1 + x^2} is not differentiable at these points. However, the question asks for ∣dydz∣\left|\frac{dy}{dz}\right|.

If ∣x∣<1|x| < 1, dydz=1\frac{dy}{dz} = 1. If ∣x∣>1|x| > 1, dydz=−1\frac{dy}{dz} = -1.

In both cases where the derivative exists (x≠1,x≠−1x \neq 1, x \neq -1), the absolute value of dydz\frac{dy}{dz} is 1. ∣dydz∣=∣1∣=1\left|\frac{dy}{dz}\right| = |1| = 1 if ∣x∣<1|x| < 1. ∣dydz∣=∣−1∣=1\left|\frac{dy}{dz}\right| = |-1| = 1 if ∣x∣>1|x| > 1.

Thus, for all xx where the derivative exists, ∣dydz∣=1\left|\frac{dy}{dz}\right| = 1.