Differentiation Question 355
Question: If $ x=\log p $ and $ y=\frac{1}{p} $ , then
Options:
A) $ \frac{d^{2}y}{dx^{2}}-2p=0 $
B) $ \frac{d^{2}y}{dx^{2}}+y=0 $
C) $ \frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}=0 $
D) $ \frac{d^{2}y}{dx^{2}}-\frac{dy}{dx}=0 $
Show Answer
Answer:
Correct Answer: C
Solution:
$ x=\log p\Rightarrow p=e^{x}\Rightarrow y={e^{-x}} $
$ \Rightarrow \frac{dy}{dx}=-{e^{-x}} $ and $ \frac{d^{2}y}{dx^{2}}={e^{-x}};\therefore \frac{d^{2}y}{dx^{2}}+\frac{dy}{dx}=0 $ .
BETA
