Differential equations

**yeloc** · September 14th 2008, 05:04 PM

Use the definition of the derivative to find f' (x).

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

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

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

**Jhevon** · September 14th 2008, 05:10 PM

Originally Posted by yeloc

Use the definition of the derivative to find f' (x).

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

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

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

**yeloc** · September 14th 2008, 05:15 PM

Originally Posted by Jhevon

it is wrong

and this is not a differential equation

How is it not when you have to use a differential equation to solve it?

**Chris L T521** · September 14th 2008, 05:17 PM

Originally Posted by yeloc

Use the definition of the derivative to find f' (x).

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

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

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

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

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

--Chris

**Jhevon** · September 14th 2008, 05:26 PM

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

**Jhevon** · September 14th 2008, 05:27 PM

Originally Posted by Chris L T521

If it was $f'(x)=-\frac{\sqrt{x}}{2x^{\color{red}2}}$ , 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, $\sqrt{x^2} = |x|$

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

**yeloc** · September 14th 2008, 05:39 PM

After I substituted $x+ \Delta x$ 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 $\Delta x$ 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?

**Jhevon** · September 14th 2008, 05:44 PM

Originally Posted by yeloc

After I substituted $x+ \Delta x$ 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 $\Delta x$ 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

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