Differentiation Question 334
Question: If $ y\sqrt{x^{2}+1}=\log { \sqrt{x^{2}+1}-x } $ , then $ (x^{2}+1)\frac{dy}{dx}+xy+1= $
[Roorkee 1978; Kurukshetra CEE 1998]
Options:
A) 0
B) 1
C) 2
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ y\sqrt{x^{2}+1}=\log { \sqrt{x^{2}+1}-x } $
Differentiating both sides w.r.t. x, we get $ \frac{dy}{dx}\sqrt{x^{2}+1}+y.\frac{1}{2\sqrt{x^{2}+1}}.2x=\frac{1}{\sqrt{x^{2}+1}-x}\times { \frac{1}{2}\frac{2x}{\sqrt{x^{2}+1}}-1 } $
Therefore $ (x^{2}+1)\frac{dy}{dx}+xy=\sqrt{x^{2}+1}.\frac{-1}{\sqrt{x^{2}+1}} $
Therefore $ (x^{2}+1)\frac{dy}{dx}+xy+1=0 $ .
BETA
