Slope of the Tangent Line-Help Please

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Find the slope of the tangent line to the curve

f(x) =

2x
----------
1 + x^2

at point x= 3


Any help would be greatly appreciated.

Thank you.
 
One way:
f(x) = 2x*(1 + x^2)^-1

d(uv) = u*d(v) + v*d(u)

u = 2x
d(u) = 2 d(x)
v=(1+x^2)^-1
d(v) = -2x(1+x^2)^-2 d(x)

f'(x) = (2x)(-2x)/(1+x^2)^2 + 2/(1+x^2) at point x= 3
 
Are you supposed to use the Quotient Formula and differentiate, or are you supposed to be working from the "limit" definition?

Please reply with the set-up you're supposed to use, if the reply Gene posted didn't answer your question. Thank you.

Eliz.
 
There's no requirement to using either(Quotient Rule or limit definition). I tried both, got stuck and different answers.

I apologize....to make it clearer its

f(x) = 2x / 1 + x^2


1st Step - find the slope of the tangent line of the curve f(x) at x=3
2nd Step - find the equation of the tangent line
3rd Step - Does f(x) have any hoziontal tangents, if so where.

I'm sure I can answer steps 2 and 3 once I get step 1.

Any help will be greatly appreciated.

Thanks.
 
Just plug x=3 into the f'(x) equation. (The correct answer is -.16)
That is the slope. Please clarify what you need help on beyond that.
 
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