# Differential equations

• Sep 14th 2008, 05:04 PM
yeloc
Differential equations
Use the definition of the derivative to find f' (x).

f (x)=$\displaystyle \frac {1}{\sqrt(x)}$

The answer I got is $\displaystyle \frac{-\sqrt(x)}{2x}$

Can someone tell me if this is right? If it isn't, I will post my work to see where I went wrong.
• Sep 14th 2008, 05:10 PM
Jhevon
Quote:

Originally Posted by yeloc
Use the definition of the derivative to find f' (x).

f (x)=$\displaystyle \frac {1}{\sqrt(x)}$

The answer I got is $\displaystyle \frac{-\sqrt(x)}{2x}$

Can someone tell me if this is right? If it isn't, I will post my work to see where I went wrong.

it is wrong

and this is not a differential equation :p
• Sep 14th 2008, 05:15 PM
yeloc
Quote:

Originally Posted by Jhevon
it is wrong

and this is not a differential equation :p

How is it not when you have to use a differential equation to solve it?
• Sep 14th 2008, 05:17 PM
Chris L T521
Quote:

Originally Posted by yeloc
Use the definition of the derivative to find f' (x).

f (x)=$\displaystyle \frac {1}{\sqrt(x)}$

The answer I got is $\displaystyle \frac{-\sqrt{x}}{2x}$

Can someone tell me if this is right? If it isn't, I will post my work to see where I went wrong.

If it was $\displaystyle f'(x)=-\frac{\sqrt{x}}{2x^{\color{red}2}}$, then you'd be correct!

What's the derivative of $\displaystyle x^{-\frac{1}{2}}$?

Recall that $\displaystyle \frac{d}{\,dx}x^n=nx^{n-1}$

--Chris
• Sep 14th 2008, 05:26 PM
Jhevon
Quote:

Originally Posted by yeloc
How is it not when you have to use a differential equation to solve it?

you are doing basic calculus. "differential equations" refers to something else. see here
• Sep 14th 2008, 05:27 PM
Jhevon
Quote:

Originally Posted by Chris L T521
If it was $\displaystyle f'(x)=-\frac{\sqrt{x}}{2x^{\color{red}2}}$, then you'd be correct!

technically yes, technically no. it would depend on whether we are considering negative x's or positive or both. remember, $\displaystyle \sqrt{x^2} = |x|$

it is better not to simplify in this case, so there is no ambiguities. just apply the chain rule, as you so rightly directed, and call it a day
• Sep 14th 2008, 05:39 PM
yeloc
After I substituted $\displaystyle x+ \Delta x$ in for x, and used the definition of the derivative of a function to set the rest of the problem up, how would I get $\displaystyle \Delta x$ out of the denominator?

I used the conjugate method, then I got a common denominator. I just simplified from there on out. Is this not right?
• Sep 14th 2008, 05:44 PM
Jhevon
Quote:

Originally Posted by yeloc
After I substituted $\displaystyle x+ \Delta x$ in for x, and used the definition of the derivative of a function to set the rest of the problem up, how would I get $\displaystyle \Delta x$ out of the denominator?

I used the conjugate method, then I got a common denominator. I just simplified from there on out. Is this not right?

you have the right idea. whether or not you did it right, i don't know