Differentiation Question 372
Question: If $ e^{y}+xy=e $ , then the value of $ \frac{d^{2}y}{dx^{2}} $ for $ x=0 $ , is
[Kurukshetra CEE 2002]
Options:
A) $ \frac{1}{e} $
B) $ \frac{1}{e^{2}} $
C) $ \frac{1}{e^{3}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ e^{y}+xy=e. $ Differentiating w.r.t. x, we get $ e^{y}\frac{dy}{dx}+y+x\frac{dy}{dx}=0 $
…..(i) Differentiating w.r.t. x, we get $ e^{y}\frac{d^{2}y}{dx^{2}}+e^{y}{{( \frac{dy}{dx} )}^{2}}+2\frac{dy}{dx}+x\frac{d^{2}y}{dx^{2}}=0 $ …..(ii) Putting $ x=0 $ in $ e^{y}+xy=e, $ we get $ y=1 $
Putting $ x=0,\ \ y=1 $ in (i), we get $ e\frac{dy}{dx}+1=0 $
Therefore $ \frac{dy}{dx}=-\frac{1}{e} $
Putting $ x=0,\ y=1,\ \frac{dy}{dx}=-\frac{1}{e} $ in (ii), we get
$ \therefore $ .
BETA
