What is (\frac{dy}{dx})^{2} equal to?
Consider the following for the two (02) items that follow : Let x=\sec~\theta-\cos~\theta and y=\sec^{4}\theta-\cos^{4}\theta.
- A. \frac{4(y^{2}+4)}{(x^{2}+4)}
- B. \frac{4(y^{2}-4)}{(x^{2}-4)}
- C. \frac{16(y^{2}+4)}{(x^{2}+4)} ✓
- D. \frac{16(y^{2}-4)}{(x^{2}-4)}
Correct Answer: C. \frac{16(y^{2}+4)}{(x^{2}+4)}
Explanation
We know x = \sec\theta - \cos\theta. Squaring and adding 4 gives x^2 + 4 = (\sec\theta + \cos\theta)^2. Similarly, y = \sec^4\theta - \cos^4\theta. Squaring and adding 4 gives y^2 + 4 = (\sec^4\theta + \cos^4\theta)^2. Differentiating with respect to \theta: \frac{dy}{d\theta} = 4\tan\theta(\sec^4\theta + \cos^4\theta) = 4\tan\theta\sqrt{y^2+4}. And \frac{dx}{d\theta} = \tan\theta(\sec\theta + \cos\theta) = \tan\theta\sqrt{x^2+4}. Thus, \frac{dy}{dx} = \frac{4\sqrt{y^2+4}}{\sqrt{x^2+4}}. Squaring both sides gives (\frac{dy}{dx})^2 = \frac{16(y^2+4)}{x^2+4}.
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