KCET · Maths · Differentiation
If \(y=\frac{\cos x}{1+\sin x}\), then
(a) \(\frac{d y}{d x}=\frac{-1}{1+\sin x}\)
(b) \(\frac{d y}{d x}=\frac{1}{1+\sin x}\)
(c) \(\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{2} \sec ^2\left(\frac{\pi}{4}-\frac{\mathrm{x}}{2}\right)\)
(d) \(\frac{d y}{d x}=\frac{1}{2} \sec ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)\)
- A Only b is correct
- B Only a is correct
- C Both a and c are correct
- D Both \(b\) and d are correct
Answer & Solution
Correct Answer
(C) Both a and c are correct
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & y=\frac{\cos x}{1+\sin x} \\ & y^{\prime}=\frac{-\sin x(1+\sin x)-\cos x(\cos x)}{(1+\sin x)^2}=\frac{-(1+\sin x)}{(1+\sin x)^2} \\ & y^{\prime}=\frac{-1}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}=\frac{-1}{1+\sin x} \\ & =\frac{-1}{2 \cdot\left(\frac{1}{\sqrt{2}} \cdot \cos \frac{x}{2}+\frac{1}{\sqrt{2}} \sin \frac{x}{2}\right)^2}=\frac{-1}{2 \cdot \cos ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)} \\ & =\frac{-1}{2} \cdot \sec ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)\end{aligned}\)
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