Need help finding the derivative of this function!

1. Apr 28, 2012

charmedbeauty

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

f(x)=x(x2-4)1/2

2. Relevant equations

3. The attempt at a solution

so I started by, let,

u=x

u'=1

v=(x2-4)1/2

v'=1/2(x2-4)-1/2(2x)

so using the product rule

f'(x)= (x2-4)1/2+x(1/2(x2-4)-1/2(2x))

=(x2-4)1/2+x2(x2-4)-1/2

Now do I just differentiate the second term again Im not quiete understanding what needs to be done here??

Last edited: Apr 28, 2012
2. Apr 28, 2012

NewtonianAlch

Where did the 4 come from?

3. Apr 28, 2012

charmedbeauty

Sorry error in the question i'll edit now.

4. Apr 28, 2012

Mentallic

Ok so you changed the question to $$f(x)=x\sqrt{x^2-4}$$ ?

You have both the square root and the power to a half. Choose one or the other, but not both :tongue:

You have a function f(x) that is comprised of the product of two other functions g(x) and h(x), so $$f(x)=g(x)h(x)$$ where g(x)=x and $h(x)=\sqrt{x^2-4}$. Now, to take the derivative of f(x) we use the rule

$$f'(x)=g(x)h'(x)+g'(x)h(x)$$

Or, if we let g(x)=u and h(x)=v,

$$f'(x)=uv'+u'v$$

So once you've found the derivative of u and v, you plug u, v, u' and v' into the formula to obtain the derivative of f.

5. Apr 28, 2012

charmedbeauty

But If you look at my working I have already done that? but I still get left with x2(x2-4)-1/2+the first term.

for starters where does that fraction even come from?

and when I differentiate once I don't get the answer I get

=(x2-4)1/2+x2(x2-4)-1/2

6. Apr 28, 2012

NewtonianAlch

That second term should be x^2/(....) not x^2(...)

That answer is the same, they've just rationalised. Multiply through by the denominator of the fraction, and you will get the answer they gave. It's the same thing.

7. Apr 28, 2012

charmedbeauty

Ok Thanks!

8. Apr 28, 2012

Mentallic

Yes, I know. I was trying to help you get rid of your uncertainties as to what you need to do from there

9. Apr 28, 2012

charmedbeauty

OOOHHH thanks a bunch mentallic!!!:tongue2: