# Thread: Intro to Derivatives - Quotient Rule

1. ## Intro to Derivatives - Quotient Rule

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:

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

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

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

Now what?

2. 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)?

3. 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: $\displaystyle f(x) = ax^2 + bx + c$

$\displaystyle f'(x) = \frac {(x+4)(x-5)}{(x-2)^2}$

4. .. I'm assuming you took the derivative right