Question

If \(x=\sec\theta+\cos\theta,\ y=\sec^n\theta+\cos^n\theta\), then \(\frac{dy}{dx}\) is equal to

• A. \(\frac{n^2(y^2+4)}{x^2+4}\)
• B. \(\frac{n^2(y^2-4)}{x^2-4}\)
• C. \(n\left(\sqrt{\frac{y^2+4}{x^2+4}}\right)\)
• D. \(n\left(\sqrt{\frac{y^2-4}{x^2-4}}\right)\)

Hint:

Show \(x^2-4=(\sec\theta-\cos\theta)^2\) and \(y^2-4=(\sec^n\theta-\cos^n\theta)^2\).

Answer 1

The Correct Option is D

Step 1: Differentiate x.
\[ \frac{dx}{d\theta} = \sec\theta\tan\theta - \sin\theta \] Write \(\sin\theta = \tan\theta\cos\theta\). Then \(\frac{dx}{d\theta} = \tan\theta(\sec\theta-\cos\theta)\).

Step 2: Differentiate y.
\[ \frac{dy}{d\theta} = n\sec^{n}\theta\tan\theta - n\cos^{n-1}\theta\sin\theta \] Using \(\sin\theta=\tan\theta\cos\theta\) again, \(\frac{dy}{d\theta} = n\tan\theta(\sec^n\theta-\cos^n\theta)\).

Step 3: Divide.
\[ \frac{dy}{dx} = \frac{n(\sec^n\theta-\cos^n\theta)}{\sec\theta-\cos\theta} \]

Step 4: Express in x and y.
Now \(x^2 = \sec^2\theta+2+\cos^2\theta\), so \(x^2-4 = (\sec\theta-\cos\theta)^2\). In the same way \(y^2-4 = (\sec^n\theta-\cos^n\theta)^2\), because \(\sec^n\theta\cos^n\theta=1\).
\[ \frac{dy}{dx} = n\sqrt{\frac{y^2-4}{x^2-4}} \]

Step 5: Check the options.
Options 1 and 3 have \(+4\), which do not match the squares above. Option 2 has an extra square and no root. Option 4 matches.

Final Answer:
The derivative is \(n\sqrt{\frac{y^2-4}{x^2-4}}\), option 4. \[ \boxed{n\sqrt{\dfrac{y^2-4}{x^2-4}}} \]