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March 6th, 2018, 08:25 AM  #1 
Newbie Joined: Nov 2016 From: Croatia Posts: 4 Thanks: 0  Trying to understand integration with very basic knowledge of derivation
I have to show that this equation is true, I know I have to find the derivative of the right side, but I don't know how to find derivative of the square root.

March 6th, 2018, 11:50 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,970 Thanks: 807 
Do you know the power rule? The derivative of $x^n$ is $nx^{n1}$. $\sqrt{x}= x^{1/2}$ so the derivative is $\frac{1}{2}x^{1/2}= \frac{1}{2\sqrt{x}}$. Further, the derivative of ln(x) is $\frac{1}{x}$. Using the "chain rule", the derivative of $ln\left(x+ \sqrt{x^2+ 1}\right)$ is $\frac{1}{x+ \sqrt{x^2+ 1}}$ times the derivative of $x+ \sqrt{x^2+ 1}= x+ (x^2+ 1)^{1/2}$. And that derivative is $1+ (1/2)(x^2+1)^{1/2}(2x)$$= 1+ \frac{x}{\sqrt{x^2+ 1}}$. So the derivative you seek is $\frac{1+ \frac{x}{\sqrt{x^2+ 1}}}{x+ \sqrt{x^2+ 1}}$ Multiplying numerator and denominator by $\sqrt{x^2+ 1}$ simplifies it slightly to $\frac{x+ \sqrt{x^2+ 1}}{x\sqrt{x^2+ 1}+ 1}$ Last edited by greg1313; March 6th, 2018 at 12:36 PM. 
March 6th, 2018, 12:41 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,767 Thanks: 1017 Math Focus: Elementary mathematics and beyond 
$$\frac{1+ \frac{x}{\sqrt{x^2+ 1}}}{x+ \sqrt{x^2+ 1}}=\frac{\frac{\sqrt{x^2+1}+x}{ \sqrt{x^2+1}}}{x+\sqrt{x^2+1}}=\frac{1}{\sqrt{x^2+ 1}}$$


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