My Math Forum Calculate F'(x)
 User Name Remember Me? Password

 Calculus Calculus Math Forum

 January 17th, 2010, 03:51 PM #1 Senior Member   Joined: Nov 2009 Posts: 169 Thanks: 0 Calculate F'(x) F(x) = $\int _1\,^{cosx}\!\sqrt {1-t^2}dt$ why can't I just sub cosx in and get $\sqrt{1-cos^2 x}$ ?
January 17th, 2010, 05:58 PM   #2
Global Moderator

Joined: May 2007

Posts: 6,806
Thanks: 716

Re: Calculate F'(x)

Quote:
 Originally Posted by 450081592 F(x) = $\int _1\,^{cosx}\!\sqrt {1-t^2}dt$ why can't I just sub cosx in and get $\sqrt{1-cos^2 x}$ ?
Back to basics. You want the x derivative of something of the form f(u(x)).
From elem. calculus you have df/dx = (df/du)(du/dx).
For your problem, you have u=cosx.

$\sqrt{1-cos^2 x}$ is the df/du part. du/dx = -sinx is needed to get the correct result.

 January 18th, 2010, 03:52 PM #3 Newbie   Joined: Jan 2010 Posts: 1 Thanks: 0 Re: Calculate F'(x) If $F(x)=\int_{a}^{g(x)}f(t)dt$ then: $F'(x)=f(g(x)) g#39;(x)$ where a is a constant.

 Tags calculate

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post sajjad mt Calculus 12 December 2nd, 2013 08:25 AM Chikis Algebra 4 June 13th, 2013 09:03 PM pitrpan Advanced Statistics 2 April 14th, 2011 01:09 PM stole Number Theory 3 August 15th, 2009 06:41 AM Espanol Algebra 3 April 2nd, 2009 10:03 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top