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greenieleenie Offline
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Can anyone help me right at this moment with derivatives (beginner stuff)? - May 20th 2009, 04:21 AM

I am really having a bit of difficulty with product rule with x's.

For example:
f(x)=(2x)/√(4x-3)

The answer is f'(x)=(4x-6)/(4x-3)^(3/2)

I just can't get to that answer! Can someone please help me and do a quick tutor session? Thanks.
   
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Re: Can anyone help me right at this moment with derivatives (beginner stuff)? - May 20th 2009, 04:30 AM

I cant help with that but, go to google and search alaska free tutors and there will be someone who can! there is a really cool website that is run by our library system.


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Re: Can anyone help me right at this moment with derivatives (beginner stuff)? - May 24th 2009, 02:38 AM

g(x)f'(x) - f(x)g'(x) / (g(x))^2

f(x)=(2x)/√(4x-3)

Think of (2x) as f(x)
√(4x-3) as g(x)
   
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Re: Can anyone help me right at this moment with derivatives (beginner stuff)? - May 24th 2009, 01:44 PM

This one might be easier using the quotient rule but if you want product rule ok..

f = 2x(4x-3)^(-1/2)

Let u = 2x and u' = 2, v = (4x-3)^(-1/2) and v' = -2(4x-3)^(-3/2).

We have f(x) = u(x)v(x) and therefore f' = uv' + u'v,

so f' = -4x(4x-3)^(-3/2) + 2(4x-3)^(-1/2), and it's just an algebra problem to get it into the form that you gave as your answer.

Multiplying the last term by (4x-3) we get
f' = -4x(4x-3)^(-3/2) + 2(4x-3)(4x-3)^(-3/2)

And we have like denominators so our numerator is
-4x + 2(4x-3) = -4x + 8x - 6 = 4x - 6

And our answer
f'(x) = (4x-6)/(4x-3)^(3/2)
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