# [SOLVED] Derivative Quotient Rule

• October 20th 2009, 12:52 PM
rawkstar
[SOLVED] Derivative Quotient Rule
I'm having trouble using the quotient rule of the derivative (i know how to do the power rule though). I'm not sure how I simplify in order to get the final answer/derivative. I need to use it in order to solve my question.

i need to determine the points at which the graph of the function
f(x)= (x^2)/(x-1) has a horizontal tangent.

what i think that i have to do is find the derivative of the function. then find the values of x by letting the derivative equal 0. then find the y values of the new x values by plugging them into the original equation. Am I right?
• October 20th 2009, 01:02 PM
Quote:

Originally Posted by rawkstar
I'm having trouble using the quotient rule of the derivative (i know how to do the power rule though). I'm not sure how I simplify in order to get the final answer/derivative. I need to use it in order to solve my question.

i need to determine the points at which the graph of the function
f(x)= (x^2)/(x-1) has a horizontal tangent.

what i think that i have to do is find the derivative of the function. then find the values of x by letting the derivative equal 0. then find the y values of the new x values by plugging them into the original equation. Am I right?

The horizontal tangent is given by finding $\lim_{x->\infty}\frac{x^2}{x-1}$
• October 20th 2009, 01:04 PM
Or wait! no no no, forget that. lol, I was thinking asymptote., sorry. I'll get to the real question in a minute.
• October 20th 2009, 01:07 PM
Quote:

Originally Posted by rawkstar
I'm having trouble using the quotient rule of the derivative (i know how to do the power rule though). I'm not sure how I simplify in order to get the final answer/derivative. I need to use it in order to solve my question.

i need to determine the points at which the graph of the function
f(x)= (x^2)/(x-1) has a horizontal tangent.

what i think that i have to do is find the derivative of the function. then find the values of x by letting the derivative equal 0. then find the y values of the new x values by plugging them into the original equation. Am I right?

Yes, find the derivative and set it to zero. The quotient rule looks like this:

$\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{g(x)f'(x)-g'(x)f(x)}{(g(x))^2}$

Do you understand how to apply this formula? I know it's difficult to remember at first, but you'll get use to it. If you have trouble applying it, I'll show you how. But see if you can do it first.
• October 20th 2009, 01:31 PM
rawkstar
i know how to apply it
it starts as (and sorry i don't know how to use the html thing)

((x-1)(2x)-(x^2)(1))/(x-1)^2
• October 20th 2009, 01:43 PM
rawkstar
oh i think i got it now
can someone check my work?

the derivative would be (x^2-2x)/(x^2-2x+1)

so 0=(x^2-2x)/(x^2-2x+1)
x=0 and x=2

then f(0)=0 and f(2)=4

so the points would be (0,0) and (2,4)
• October 20th 2009, 02:37 PM
Quote:

Originally Posted by rawkstar
oh i think i got it now
can someone check my work?

the derivative would be (x^2-2x)/(x^2-2x+1)

so 0=(x^2-2x)/(x^2-2x+1)
x=0 and x=2

then f(0)=0 and f(2)=4

so the points would be (0,0) and (2,4)

Yes, this is correct.
• October 20th 2009, 03:09 PM
rawkstar
Thank you so much