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What is the value of \(\frac{d}{dx}(\frac{cos⁡x}{sec⁡x \,tan⁡x})\)?(a) -tan^2⁡ x(3 sin^2⁡ x – cos^2 x)(b) -cot^2 x(3 sin^2⁡ x – cos^2 x)(c) -cot^2 x(3 sin^2⁡ x + cos^2 x)(d) -tan^2⁡ x(3 sin^2⁡ x + cos^2 x)I have been asked this question in unit test.Question is from Derivatives in chapter Limits and Derivatives of Mathematics – Class 11

Answer»

Right choice is (c) -cot^2 x(3 sin^2⁡ x + cos^2 x)

To explain: \(\frac{d}{dx}(\frac{cos⁡x}{sec⁡x \,tan⁡x}) = \frac{d}{dx}(\frac{cos⁡x}{\frac{1}{cos⁡x}\frac{sin⁡x}{cos⁡x}})\)

=\(\frac{d}{dx}(\frac{cos^3⁡x}{sin⁡x})\)

Using quotient rule, we KNOW that, \(\frac{d}{dx} (\frac{F}{G}) = \frac{g.\frac{d}{dx} (f) – f.\frac{d}{dx}(g)}{g^2}\)

Here, f = cos^3⁡ x and g = sin ⁡x

\(\frac{d}{dx}(\frac{cos^3⁡x}{sin⁡x})\) = \(\frac{sin⁡x.\frac{d}{dx} (cos^3⁡x)-cos^3⁡x.\frac{d}{dx}(sinx)}{sin^2⁡x}\)

\(\frac{d}{dx}(\frac{cos^3⁡x}{sin⁡x})\) = \(\frac{sin⁡x (-3 cos^2⁡ x \,sin⁡ x) – cos^3⁡x(cos⁡x)}{sin^2⁡x}\)

\(\frac{d}{dx}(\frac{cos^3⁡x}{sin⁡x})\) = \(\frac{-3 sin^2⁡x \,cos^2⁡x – cos^4⁡x}{sin^2⁡x}\)

\(\frac{d}{dx}(\frac{cos^3⁡x}{sin⁡x})\) = -cot^2⁡ x(3 sin^2⁡ x + cos^2 x)


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