Differentiation Question 260
Question: If $ y=\frac{e^{2x}+{e^{-2x}}}{e^{2x}-{e^{-2x}}} $ , then $ \frac{dy}{dx}= $
[AI CBSE 1988]
Options:
A) $ \frac{-8}{{{(e^{2x}-{e^{-2x}})}^{2}}} $
B) $ \frac{8}{{{(e^{2x}-{e^{-2x}})}^{2}}} $
C) $ \frac{-4}{{{(e^{2x}-{e^{-2x}})}^{2}}} $
D) $ \frac{4}{{{(e^{2x}-{e^{-2x}})}^{2}}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ y=\frac{e^{2x}+{e^{-2x}}}{e^{2x}-{e^{-2x}}} $
$ \therefore \frac{dy}{dx}=\frac{(e^{2x}-{e^{-2x}})2(e^{2x}-{e^{-2x}})-(e^{2x}+{e^{-2x}})2(e^{2x}+{e^{-2x}})}{{{(e^{2x}-{e^{-2x}})}^{2}}} $
$ =\frac{-8}{{{(e^{2x}-{e^{-2x}})}^{2}}} $ .
BETA
