Finding the derivative of a function

1. Dec 23, 2008

meeklobraca

1. The problem statement, all variables and given/known data

y=(5-x2)sqrt(x)

2. Relevant equations

3. The attempt at a solution

I got as far as

(5-x2)(1/2)(x)^-1/2 + (x)^1/2 (-x)(-2x)

Im using the chain rule here I believe but im a tad lost as where to go from here, if ive even done it correctly up to this point. Any help would be greatly appreciated.

Thanks!

2. Dec 23, 2008

Vagrant

How do you get this? The derivative of x2 is 2x. So it should be $$\sqrt{x}$$(-2x).

3. Dec 23, 2008

meeklobraca

Im not sure, this chain rule is explained extremely poorly in my manual. Isnt the equatrion I should be using

u * dv/dx + v * du/dx ?

4. Dec 23, 2008

Vagrant

Yes.
u=(5-x2)
v=$$\sqrt{x}$$
Now u dv/dx = (5-x2) ( 1/2 x^-1/2), which you've got,
and v du/dx = $$\sqrt{x}$$ (-2x).

Add the two parts to get dy/dx.

5. Dec 23, 2008

meeklobraca

I dont know where to go from there.

I got (1/2x^-1/2)(5-x2)

But that seems wrong.

6. Dec 23, 2008

Vagrant

That's just one part. Add u dv/dx and v du/dx to get dy/dx. Then simplify the expression.

7. Dec 23, 2008

meeklobraca

Thats what I thought I did. I had

1/2x^-1/2 (5 - x2 + 2x - 2x) which led me to my answer.

8. Dec 23, 2008

Vagrant

How did you get this?
I get,
dy/dx = (5-x2)(1/(2$$\sqrt{x}$$)) - 2x3/2
dy/dx = 5/(2$$\sqrt{x}$$) - 1/2x3/2 -2x3/2

Simplify.

9. Dec 23, 2008

meeklobraca

I dont get how you get to that second line of dy/dx. how did 2 factors minus 1 factor = 3 factors all subtracted from each other. In fact, how you got -2x^3/2 isnt clear either.

Last edited: Dec 23, 2008
10. Dec 24, 2008

HallsofIvy

Staff Emeritus
The "distributive law": a(b+ c)= ab+ bc.

$$(5- x^2)(1/(2\sqrt{x})= 5(1/(2\sqrt{x})- x^2(1/2\sqrt{x})$$

Further, $\sqrt{x}= x^{1/2}$ so $1/(2/sqrt{x})= (1/2)x^{-1/2}$ and $x^2(1/2\sqrt{x})= (1/2)x^{2- 1/2}= (1/2)x^{3/2}$.

11. Dec 24, 2008

meeklobraca

We havent been taught that method, I imagine I wont get marks for it if I do it that way. I have to use one of the derivative rules to get the final answer.

12. Dec 24, 2008

jgens

Which method are you reffering to? Halls' use of the distributive property or the use of the product rule (d(uv)/dx = u'v + v'u) and power rule (d(x^n) = nx^(n-1))?

13. Dec 24, 2008

meeklobraca

The distributive property

14. Dec 24, 2008

jgens

If you've had an elementary algebra course you should be familiar with the distributive property. I don't think you would lose marks for applying an algebraic property to simplify your answers.

15. Dec 24, 2008

meeklobraca

From what ive gathered from the course I think I would. The professor wants an answer to the question using the rules in the unit. There must be an answer to the question using the deriviative rules isnt there?

16. Dec 24, 2008

jgens

Yes, but it's not in it's simplest form. I would clarify with your professor that he doesn't want you to apply algebraic rules to simplify expressions.

17. Dec 24, 2008

HallsofIvy

Staff Emeritus
It was not a question of "derivative rules", it is a question of using algebra. I can't believe that you are not "allowed" to use what you have learned before this course. You are allowed to use arithmetic aren't you?