Differentiation Question 380
Question: If $ y={{( x+\sqrt{1+x^{2}} )}^{n}}, $ then $ (1+x^{2})\frac{d^{2}y}{dx^{2}}+x\frac{dy}{dx} $ is
[AIEEE 2002]
Options:
A) $ n^{2}y $
B) $ -n^{2}y $
C) $ -y $
D) $ 2x^{2}y $
Show Answer
Answer:
Correct Answer: A
Solution:
$ y={{(x+\sqrt{1+x^{2}})}^{n}}\Rightarrow \frac{dy}{dx}=n{{(x+\sqrt{1+x^{2}})}^{n-1}}( 1+\frac{x}{\sqrt{1+x^{2}}} ) $
Therefore $ \frac{dy}{dx}=\frac{n{{(x+\sqrt{1+x^{2}})}^{n}}}{\sqrt{1+x^{2}}} $
Therefore $ (\sqrt{1+x^{2}})\frac{dy}{dx}=n{{(x+\sqrt{1+x^{2}})}^{n}} $
Therefore $ \frac{d^{2}y}{dx^{2}}.\sqrt{1+x^{2}}+\frac{dy}{dx}( \frac{x}{\sqrt{1+x^{2}}} ) $
$ =n^{2}{{( x+\sqrt{1+x^{2}} )}^{n-1}}( 1+\frac{x}{\sqrt{1+x^{2}}} ) $
Therefore $ (1+x^{2}).\frac{d^{2}y}{dx^{2}}+x.\frac{dy}{dx}=n^{2}{{(x+\sqrt{1+x^{2}})}^{n}} $
Therefore $ (1+x^{2})\frac{d^{2}y}{dx^{2}}+x.\frac{dy}{dx}=n^{2}y $ .
BETA
