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March 2nd, 2016, 07:18 PM  #1 
Member Joined: May 2015 From: Earth Posts: 64 Thanks: 0  Confusion over the Fundamental Theorem
So I've been looking at ways to rewrite arcsin, and I found this: arcsin (x/a) + C = the integral of 1 over the square root of a^2 minus x^2 dx This is correct But when I try to graph the DERIVATIVE of arcsin (x/a), I expected it to be the same graph as 1 over the square root of a^2 minus x^2 dx, which is the same thing as above but without the integral. However, this does not work, instead the a^2 must be a 1 for them to be the same. Why is this? 
March 2nd, 2016, 07:39 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,511 Thanks: 2514 Math Focus: Mainly analysis and algebra 
Are you graphing it correctly, or is your derivative incorrect? ${\mathrm d \over \mathrm d x}\arcsin {x \over a} = \frac1a {1 \over \sqrt{1 \left({x \over a}\right)^2}}$ 
March 2nd, 2016, 07:45 PM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 2,780 Thanks: 1431 
What value did you use for the constant, $a$? I used $a=3$, and graphed both the derivative of $\arcsin(x/3)$ and $\dfrac{1}{\sqrt{9x^2}}$. Note that I added 1 to the square root function, effectively shifting it up one unit so it could be seen ... without doing so, the two graphs were superimposed. Last edited by skipjack; March 3rd, 2016 at 02:01 AM. 
March 2nd, 2016, 09:48 PM  #4 
Member Joined: May 2015 From: Earth Posts: 64 Thanks: 0 
I see what I've done wrong, sorry guys, thanks for always offering help though!!


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