logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

If \( y=\log [x+\sqrt{x^2+1}],\) prove that \( (x^2+1)\large\frac{d^2y}{dx^2}+x\frac{dy}{dx}=0.\)

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $\large\frac{d}{dx}$$(\log x)=\large\frac{1}{x}$
Step 1:
$y=\log(x+\sqrt{x^2+1})$
$\Rightarrow \large\frac{dy}{dx}=\bigg[\large\frac{1}{x+\sqrt{x^2+1}}.$$(1+\large\frac{1}{2\sqrt{x^2+1}.2x}\bigg]$
$\Rightarrow \large\frac{1}{x+\sqrt{x^2+1}}\bigg[\large\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}\bigg]$
$\Rightarrow \large\frac{1}{\sqrt{x^2+1}}$
Step 2:
$\large\frac{d^2y}{dx^2}=\large\frac{0-1.1/2\sqrt{x^2+1}.2x}{(\sqrt{x^2+1)^2}}$
$\Rightarrow \large\frac{-x}{(x^2+1)\sqrt{x^2+1}}$
$\Rightarrow (x^2+1)\large\frac{d^2y}{dx^2}=\frac{-x}{\sqrt{x^2+1}}$
$\therefore (x^2+1)\large\frac{d^2y}{dx^2}$$+x\large\frac{dy}{dx}=\frac{-x}{\sqrt{x^2+1}}+\frac{x}{\sqrt{x^2+1}}$
$\Rightarrow 0$
Hence proved.
answered Nov 15, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...