# Derive this?

• Jan 23rd 2011, 11:25 AM
Vamz
Derive this?
$\displaystyle f(x)=\displaystyle\sqrt{49-x^2}+7 \cos^{-1}\left(\frac{x}{7}\right)$

So that F'(x) has this format:

$\displaystyle \displaystyle -\left(\frac{x+c}{mx+n}\right)^{p}$

When I derive this, I get:
$\displaystyle \diplaystyle\frac{-x}{\sqrt{(49-x^2})}+\frac{1}{\sqrt{(1-(\frac{x}{7})^2)}}$

How do I get in into that format?
• Jan 23rd 2011, 11:37 AM
skeeter
Quote:

Originally Posted by Vamz
$\displaystyle f(x)=\displaystyle\sqrt{49-x^2}+7 \cos^{-1}\left(\frac{x}{7}\right)$

So that F'(x) has this format:

$\displaystyle \displaystyle -\left(\frac{x+c}{mx+n}\right)^{p}$

When I derive this, I get:
$\displaystyle \diplaystyle\frac{-x}{\sqrt{(49-x^2})}+\frac{1}{\sqrt{(1-(\frac{x}{7})^2)}}$

How do I get in into that format?

$\displaystyle \diplaystyle \frac{-x}{\sqrt{49-x^2}}- \frac{7}{7} \cdot \frac{1}{\sqrt{1-(\frac{x}{7})^2}}$

$\displaystyle \diplaystyle \frac{-x}{\sqrt{49-x^2}}- \frac{7}{\sqrt{49} \cdot \sqrt{1-(\frac{x}{7})^2}}$

$\displaystyle \diplaystyle \frac{-x}{\sqrt{49-x^2}}- \frac{7}{\sqrt{49-x^2}}$

$\displaystyle \diplaystyle \frac{-(x+7)}{\sqrt{49-x^2}}$
• Jan 23rd 2011, 11:41 AM
SENTINEL4
You get the derivative $\displaystyle -\frac{x}{\sqrt{49-x^2}}-\frac{1}{\sqrt{1-\frac{x^2}{49}}}$ = $\displaystyle -\frac{x}{\sqrt{49-x^2}}-\frac{1}{\sqrt{\frac{49-x^2}{49}}}$ = $\displaystyle -\frac{x}{\sqrt{49-x^2}}-\frac{1}{\frac{\sqrt{49-x^2}}{7}}$

From there by doing some simple things you can get the form you want
• Jan 23rd 2011, 11:44 AM
Vamz
Thanks!
I don't think that is quite right however!

There needs to be a "p" value & it cannot be "1"
• Jan 23rd 2011, 11:47 AM
SENTINEL4
If you do some things more you will get a value for p that isn't 1
• Jan 23rd 2011, 11:48 AM
Vamz
• Jan 23rd 2011, 11:56 AM
SENTINEL4
$\displaystyle -\frac{x}{\sqrt{49-x^2}}-\frac{7}{\sqrt{49-x^2}}$ = $\displaystyle -\frac{x+7}{\sqrt{49-x^2}}$ = $\displaystyle -\frac{x+7}{\sqrt{7-x}\sqrt{7+x}}$

one-two more steps and you'll get what you want
• Jan 23rd 2011, 12:05 PM
Aryth
Quote:

Originally Posted by Vamz
$\displaystyle f(x)=\displaystyle\sqrt{49-x^2}+7 \cos^{-1}\left(\frac{x}{7}\right)$

So that F'(x) has this format:

$\displaystyle \displaystyle -\left(\frac{x+c}{mx+n}\right)^{p}$

When I derive this, I get:
$\displaystyle \diplaystyle\frac{-x}{\sqrt{(49-x^2})}+\frac{1}{\sqrt{(1-(\frac{x}{7})^2)}}$

How do I get in into that format?

You're derivative is equivalent to what they have except at one point. The derivative of the inverse cosine has a negative on it, not a positive, turning your answer to:

$\displaystyle \frac{-x}{\sqrt{49 - x^2}} + \frac{-1}{\sqrt{1 - \left(\frac{x}{7}\right)^2}}$

Then you take the second term and multiply it by $\displaystyle 1 = \frac{7}{7}$ to get:

$\displaystyle -\frac{x}{\sqrt{49-x^2}}-\frac{7}{\sqrt{49-x^2}}$

Then you put everything together to get:

$\displaystyle -\frac{x+7}{\sqrt{49-x^2}}$

We know that the thing inside the square root can factor:

$\displaystyle -\frac{x + 7}{\sqrt{7-x}\sqrt{7+x}}$

From there, you simply recognize that:

$\displaystyle \frac{x}{\sqrt{x}} = \sqrt{x}$

To get:

$\displaystyle -\frac{\sqrt{7 + x}}{\sqrt{7 - x}} = -\sqrt{\frac{7+x}{7-x}}$

This will then give:

$\displaystyle -\left(\frac{7+x}{7-x}\right)^{\frac{1}{2}}$

So, the values are:

c = 7
m = -1
n = 7
p = 1/2