Find the derivative then make it into the format:

I got it to the form

Soooo shouldn't..

c=7

m=-1

n=49

p=1

Or have I done something wrong?

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- Jan 28th 2010, 08:35 PMLolcatsDerivative of a inverse trig function... Is my answer wrong?

Find the derivative then make it into the format:

I got it to the form

Soooo shouldn't..

c=7

m=-1

n=49

p=1

Or have I done something wrong? - Jan 28th 2010, 08:46 PMCalculus26
Not quite right --note you need the same power top and bottom

Note 49 -x^2 = (7-x) (7+x)

you have - [sqrt(x+7)/sqrt((7-x)] = - [(x+7)/(-x+7)]^(1/2)

p =1/2

n = 7

c=7

m = -1 - Jan 28th 2010, 08:57 PMLolcats
Wouldn't I then currently have

-(x+7)/((7-x) (7+x))^(1/2)?

I think I know where your going but I don't see it :S, Can you explain that? - Jan 28th 2010, 09:47 PMCalculus26
x+7 = 7 + x

so (x+7)/[sqrt(7+x)sqrt(7-x)]=( 7 +x)/[sqrt(7+x)sqrt(7-x)]

= sqrt(7+x)/sqrt(7-x) - Jan 28th 2010, 10:32 PMLolcats
OHh okay, thank you!