# Thread: Using the difference quotient?

1. ## Using the difference quotient?

Hi! I'm new to Math Help Forums, so you'll have to forgive me if I don't do things correctly. I'm not quite sure how to post math icons either =/

So, I just have a basic difference quotient problem that was actually included in my review packet for the beginning of calculus one.

Question:
Evaluate the function at the given value of the independent variable.

F(x) = x^3
f(x+deltax)-f(x)/delta x

I got as far as plugging it in:
(x+delta x)^3-x^2/delta x

And, I know I'm supposed to simplify things out to get rid of the delta x on the bottom. I tried multiplying out the top and I got [x^2+x(delta x) +x(delta x) + (delta x)^2][x+delta x]/delta x. I'm not entirely sure if that's right, because the delta x's kind of confuse me. I wasn't quite sure how to continue.

Thanks!

EDIT: I forgot to mention! The back of the book states that the answer is 3x^2 + 3x(delta x)+(delta x)^2, where delta x does not equal zero. Thanks again!

2. It's mostly just an algebra problem in the numerator. Notice that

$(a+b)^3 = a^3 + 3a^2 b + 3a b^2 + b^3$

so:

$\frac{f(x+ \Delta x) - f(x)}{\Delta x} = \frac{(x+ \Delta x)^3 - x^3}{\Delta x} = \frac{x^3 + 3 x^2 \Delta x + 3 x (\Delta x)^2 + (\Delta x)^3 - x^3}{\Delta x}$ $= \frac{3 x^2 \Delta x + 3 x (\Delta x)^2 + (\Delta x)^3}{\Delta x}$

If the $\Delta x$ terms are throwing you off, you can use $h$ instead. This is actually more common notation anyway.

$\frac{f(x+ h) - f(x)}{h} = \frac{(x+ h)^3 - x^3}{h} = \frac{x^3 + 3 x^2 h + 3 x h^2 + h^3 - x^3}{h}$ $= \frac{3 x^2 h + 3 x h^2 + h^3}{h}$

Does this help?

3. Oh my gosh! I couldn't believe I forgot the difference of cubes! Thank you so much, that definitely makes more sense. I was trying to multiply everything out by hand instead of using a simple equation. Thank you!