# Derivative of a inverse trig function... Is my answer wrong?

• Jan 28th 2010, 07:35 PM
Lolcats
Derivative of a inverse trig function... Is my answer wrong?
$\displaystyle f(x)=\displaystyle\sqrt{49-x^2}+7 \cos^{-1}\left(\frac{x}{7}\right)$

Find the derivative then make it into the format:
$\displaystyle \displaystyle -\left(\frac{x+c}{mx+n}\right)^{p}$

I got it to the form
$\displaystyle \frac{-(x+7)}{\sqrt{49-x^2}}$

Soooo shouldn't..
c=7
m=-1
n=49
p=1

Or have I done something wrong?
• Jan 28th 2010, 07:46 PM
Calculus26
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, 07:57 PM
Lolcats
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, 08:47 PM
Calculus26
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, 09:32 PM
Lolcats
OHh okay, thank you!