Finding the derivative of a function

[meeklobraca]

Homework Statement

y=(5-x2)sqrt(x)

Homework Equations

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 I am a tad lost as where to go from here, if I've even done it correctly up to this point. Any help would be greatly appreciated.

Thanks!

 

[Vagrant]

(x)^1/2 (-x)(-2x)

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

 

[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 ?

 

[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.

 

[meeklobraca]

I don't know where to go from there.

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

But that seems wrong.

 

[Vagrant]

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

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

 

[meeklobraca]

Thats what I thought I did. I had

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

 

[Vagrant]

2x - 2x

How did you get this?
I get,
dy/dx = (5-x²)(1/(2\sqrt{x})) - 2x^3/2
dy/dx = 5/(2\sqrt{x}) - 1/2x^3/2 -2x^3/2

Simplify.

 

[meeklobraca]

I don't 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 isn't clear either.

 

[HallsofIvy]

How did you get this?
I get,
dy/dx = (5-x²)(1/(2\sqrt{x})) - 2x^3/2
dy/dx = 5/(2\sqrt{x}) - 1/2x^3/2 -2x^3/2

Simplify.

I don't 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 isn't clear either.

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}.

 

[meeklobraca]

We haven't been taught that method, I imagine I won't get marks for it if I do it that way. I have to use one of the derivative rules to get the final answer.

 

[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))?

 

[meeklobraca]

The distributive property

 

[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.

 

[meeklobraca]

From what I've 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 isn't there?

 

[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.

 

[HallsofIvy]

From what I've 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 isn't there?

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?