COMEDK · Maths · 26. Differentiation
If \(y=|\cos x|+|\sin x|\), then \(\frac{d y}{d x}\) at \(x=\frac{2 \pi}{3}\) is
- A \(\frac{1-\sqrt{3}}{2}\)
- B 0
- C \(\frac{1}{2}(\sqrt{3}-1)\)
- D None of these
Answer & Solution
Correct Answer
(C) \(\frac{1}{2}(\sqrt{3}-1)\)
Step-by-step Solution
Detailed explanation
We have, \(y=|\cos x|+|\sin x|\) At \(x=\frac{2 \pi}{3}, \cos x\) is negative and \(\sin x\) is positive. \(\therefore \quad y=-\cos x+\sin x\) \(\Rightarrow \frac{d y}{d x}=\sin x+\cos x\)…
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