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March 6th, 2018, 08:25 AM   #1
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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.
Attached Images Capture.PNG (2.9 KB, 3 views) March 6th, 2018, 11:50 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Do you know the power rule? The derivative of $x^n$ is $nx^{n-1}$. $\sqrt{x}= x^{1/2}$ 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)$$= 1+ \frac{x}{\sqrt{x^2+ 1}}. So the derivative you seek is \frac{1+ \frac{x}{\sqrt{x^2+ 1}}}{x+ \sqrt{x^2+ 1}} Multiplying numerator and denominator by \sqrt{x^2+ 1} simplifies it slightly to \frac{x+ \sqrt{x^2+ 1}}{x\sqrt{x^2+ 1}+ 1} Thanks from greg1313 and Sebastian Garth Last edited by greg1313; March 6th, 2018 at 12:36 PM. March 6th, 2018, 12:41 PM #3 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond$$\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}}$\$ Tags basic, calculus, college, derivation, integration, knowledge, understand Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post 3uler Calculus 6 September 24th, 2014 04:38 AM questioner1 Calculus 1 August 14th, 2014 08:26 AM Emma.k Math Books 0 March 23rd, 2014 04:17 AM Twinndaddy Algebra 1 February 7th, 2014 07:12 AM conure Algebra 3 October 21st, 2013 11:33 AM

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