If x=\sin~\theta, then what is \frac{dy}{dx} equal to?

For the following three (03) items: Let y=f(x)=\frac{x \sin^{-1}x}{\sqrt{1-x^{2}}}+\ln\sqrt{1-x^{2}}.

  1. A. \theta~\sec~\theta
  2. B. \theta~\sec^{2}\theta
  3. C. \theta~\sec^{3}\theta
  4. D. 2~\tan~\theta+\theta~\sec^{2}\theta

Correct Answer: C. \theta~\sec^{3}\theta

Explanation

We established earlier that \frac{dy}{dx} = \frac{\sin^{-1}x}{(1-x^2)^{3/2}}. Given x = \sin\theta, we have \sin^{-1}x = \theta. The denominator transforms as (1-\sin^2\theta)^{3/2} = (\cos^2\theta)^{3/2} = \cos^3\theta. Therefore, \frac{dy}{dx} = \frac{\theta}{\cos^3\theta} = \theta \sec^3\theta.

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