# Math Help - Numerical Derivative

1. ## Numerical Derivative

If you know numerical derivative, that is what I'm finding.
If not, that's okay. I can translate.

Find the NDER of [tex]x^(3/5)

Also x^(2/3) .... but I think learning the top will be enough to grasp the concept, unless there is something else worth mentioning for this.

Thanks!

2. Originally Posted by Truthbetold
If you know numerical derivative, that is what I'm finding.
If not, that's okay. I can translate.

Find the NDER of [tex]x^(3/5)

Also x^(2/3) .... but I think learning the top will be enough to grasp the concept, unless there is something else worth mentioning for this.

Thanks!
wat do u mean by numerical derivative?
if im not wrong, are you looking for the answer $\frac{3}{5}x^{\frac{-2}{5}}$

3. if im not wrong, are you looking for the answer
\
That looks correct, but I don't know how to do this.

NUmerical derivative is this:

$\frac {f(a+.001)- f(a - .001)}{.002}$

If you don't get it, just find the derivative.

I don't know how to deal with (a + 0.001)^(3/5)

It isn't shortcuts, meaning the power rule and stuff, because they are next lesson.

Thanks!

4. Originally Posted by Truthbetold
..
NUmerical derivative is this:

$\frac {f(a+.001)- f(a - .001)}{.002}$

..
I don't know how to deal with (a + 0.001)^(3/5)

It isn't shortcuts, meaning the power rule and stuff, because they are next lesson.

Thanks!
where did $\frac {f(a+.001)- f(a - .001)}{.002}$ come from?
i just used power rule..

5. I am taught the power rule next lesson.
You asked for the numerical derivative, and that is what that is.
I need to use that to figure out the equation.
I prefer the power rule also, but I can't use it.

I need to eventually know this for the test as well.

6. Originally Posted by Truthbetold
I am taught the power rule next lesson.
You asked for the numerical derivative, and that is what that is.
I need to use that to figure out the equation.
I prefer the power rule also, but I can't use it.

I need to eventually know this for the test as well.
ok, i just dont get where a, +- 0.001, and 0.002 came from.. sorry..

7. Originally Posted by Truthbetold
If you know numerical derivative, that is what I'm finding.
If not, that's okay. I can translate.

Find the NDER of [tex]x^(3/5)

Also x^(2/3) .... but I think learning the top will be enough to grasp the concept, unless there is something else worth mentioning for this.

Thanks!
If I am correct your teacher is expecting you to approximate the value of the derivative of a function at a given point.

$f^{\prime}(x) \approx \frac{f(x + 0.001) - f(x - 0.001)}{0.002}$
where we are approximating the formula
$f^{\prime}(x) \approx \frac{f(x + \Delta h) - f(x - \Delta h)}{2 \Delta h}$
(If the 2 in the denominator looks a little funny remember that the denominator is the size of the interval over which we are applying the secant function.)

So for $f(x) = x^{3/5}$ we get
$f^{\prime}(x) \approx \frac{(x + 0.001)^{3/5} - (x - 0.001)^{3/5}}{0.002}$

This result only has value if you know the general form for the binomial theorem:
$(x + \Delta h)^n = x^n + n \Delta h x^{n - 1} + ~ ...$
(The rest of the terms don't matter as $\Delta h$ is small.)

So
$f^{\prime}(x) \approx \frac{(x + 0.001)^{3/5} - (x - 0.001)^{3/5}}{0.002} \approx$ $\frac{x^{3/5} + \frac{3}{5} \cdot 0.001x^{3/5 - 1} - x^{3/5} + \frac{3}{5} \cdot 0.001x^{3/5 - 1}}{0.002}$

$= \frac{3}{5} \cdot \frac{2 \cdot 0.001 x^{-2/5}}{2 \cdot 0.001}$

$= \frac{3}{5}x^{-2/5}$

(If anyone objects to the concept of using the approximation form of the binomial theorem, which I believe is gotten by using differential Calculus, I agree that this makes my argument rather circular. I simply couldn't find a way to turn the approximate derivative into a function, rather than something that needed to be graphed point by point.)

-Dan