# Intro to Derivatives - Quotient Rule

• May 4th 2008, 11:48 AM
Macleef
Intro to Derivatives - Quotient Rule
http://i31.tinypic.com/2qcp7r6.png

There is no answer in my textbook for this question.
Please show a step by step solution so I could follow through...

Here's what I've done so far:

$y' = \frac {5(x + 2) - 1(5x + 2)}{(x+2)^2}$

$y' = \frac {5x + 10 - 5x - 2}{(x+2)^2}$

$y' = \frac {8}{(x+2)^2}$

Now what?
• May 4th 2008, 11:51 AM
o_O
You know that derivatives essentially represent the slopes of a function. So, if you were to have tangents with a negative slope, f'(x) must be negative.

If f(x) does not have any negative tangents, that must mean f'(x) is never less than 0. Does this happen in your expression for f'(x)?
• May 4th 2008, 12:34 PM
Macleef
Quote:

Originally Posted by o_O
You know that derivatives essentially represent the slopes of a function. So, if you were to have tangents with a negative slope, f'(x) must be negative.

If f(x) does not have any negative tangents, that must mean f'(x) is never less than 0. Does this happen in your expression for f'(x)?

Yes, it doesn't have any negative tangents because the denominator will always be positive and the numerator is a constant positive value?

OT: When you're finding the x values to substitute in the original equation, do you solve for the numerator or denominator?

For example;

Original Equation: $f(x) = ax^2 + bx + c$

$f'(x) = \frac {(x+4)(x-5)}{(x-2)^2}$
• May 4th 2008, 12:35 PM
o_O
(Clapping) .. I'm assuming you took the derivative right :p