Differentiation Question 374
Question: If $ y=\sin x+e^{x}, $ then $ \frac{d^{2}x}{dy^{2}}= $
[Karnataka CET 1999; UPSEAT 2001; Kurukshetra CEE 2002]
Options:
A) $ {{(-\sin x+e^{x})}^{-1}} $
B) $ \frac{\sin x-e^{x}}{{{(\cos x+e^{x})}^{2}}} $
C) $ \frac{\sin x-e^{x}}{{{(\cos x+e^{x})}^{3}}} $
D) $ \frac{\sin x+e^{x}}{{{(\cos x+e^{x})}^{3}}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ y=\sin x+e^{x} $
Therefore $ \frac{dy}{dx}=\cos x+e^{x} $
Therefore $ \frac{dx}{dy}={{(\cos x+e^{x})}^{-1}} $
……..(i) Again, $ \frac{d^{2}x}{dy^{2}}=-{{(\cos x+e^{x})}^{-2}}(-\sin x+e^{x})\frac{dx}{dy} $ . Substituting the value of $ \frac{dx}{dy} $ from (i), $ \frac{d^{2}x}{dy^{2}}=\frac{(\sin x-e^{x})}{{{(\cos x+e^{x})}^{2}}}{{(\cos x+e^{x})}^{-1}} $
$ =\frac{\sin x-e^{x}}{{{(\cos x+e^{x})}^{3}}} $ .
BETA
