Question

If \[ y = \tan^{-1} \left( \frac{2 - 3\sin x}{3 - 2\sin x} \right), \] then find \( \frac{dy}{dx} \).

• A. \( \frac{(3 - 2\sin x)^2}{13\sin^2 x - 24\sin x + 13} \)
• B. \( \frac{-5 \cos x}{13\sin^2 x - 24\sin x + 13} \)
• C. \( \frac{5 \sin x}{13\sin^2 x - 24\sin x + 13} \)
• D. \( \frac{-5 \sin x}{13\sin^2 x - 24\sin x + 13} \)

Hint:

For differentiating inverse trigonometric functions, use quotient rule and apply the standard derivative formulas.

Answer 1

The Correct Option is B

Step 1: Differentiating using inverse trigonometric derivative Using the derivative formula: \[ \frac{d}{dx} \tan^{-1} u = \frac{u'}{1 + u^2}. \] Let \( u = \frac{2 - 3\sin x}{3 - 2\sin x} \). Differentiating using quotient rule: \[ u' = \frac{(-3\cos x)(3 - 2\sin x) - (-2\cos x)(2 - 3\sin x)}{(3 - 2\sin x)^2}. \] \[ = \frac{-9\cos x + 6\sin x \cos x + 4\cos x - 6\sin x \cos x}{(3 - 2\sin x)^2}. \] \[ = \frac{-5\cos x}{(3 - 2\sin x)^2}. \] Applying the inverse tan derivative: \[ \frac{dy}{dx} = \frac{-5 \cos x}{13\sin^2 x - 24\sin x + 13}. \]