1. ## Cubing (x+∆x)

We just started differential equations in my AP Calculus class, and the only thing I am having trouble with is $\displaystyle [(x+(\Delta)x]^3$. I tried using the binomial theorem (for cubing) but I get the wrong answer at the end. The only place I think I could be messing up is cubing $\displaystyle (x+(\Delta)x)$.

The problem:
Use the definition of the derivative to find f'(x).

f (x)= $\displaystyle x^3+x^2$

My work thus far:

f' (x)= $\displaystyle \lim_{(\Delta)x\rightarrow0}$ $\displaystyle [(x+(\Delta)x]^3$ + $\displaystyle [(x+(\Delta)x]^2$ $\displaystyle -$$\displaystyle (x^3+x^2) DIVIDED BY \displaystyle (\Delta)x =\displaystyle x^3+x^2(\Delta)x+x(\Delta)x^2+(\Delta)x^3+x^2+2x(\ Delta)x+(\Delta)x^2-x^3-x^2 DIVIDED BY \displaystyle (\Delta)x =\displaystyle (\Delta)x[x^2+x(\Delta)x+(\Delta)x^2+2x+(\Delta)x] DIVIDED BY \displaystyle (\Delta)x The \displaystyle (\Delta)x will cancel leaving: \displaystyle [x^2+x(\Delta)x+(\Delta)x^2+2x+(\Delta)x] I plugged 0 in for \displaystyle (\Delta)x and was left with \displaystyle x^2+2x But I know this is the wrong answer (the right answer should be \displaystyle 3x^2+2x) **To clarify, DIVIDED BY \displaystyle (\Delta)x means everything on that line is divided by \displaystyle (\Delta)x. I can't figure out how to write \displaystyle (\Delta)x without the parentheses being around Delta. Nor can I figure out how to get \displaystyle (\Delta)x in the denominator of the whole expression. I tried though. 2. Originally Posted by yeloc We just started differential equations in my AP Calculus class, and the only thing I am having trouble with is \displaystyle [(x+(\Delta)x]^3. I tried using the binomial theorem (for cubing) but I get the wrong answer at the end. The only place I think I could be messing up is cubing \displaystyle (x+(\Delta)x). The problem: Use the definition of the derivative to find f'(x). f (x)= \displaystyle x^3+x^2 My work thus far: f' (x)= \displaystyle \lim_{(\Delta)x\rightarrow0} \displaystyle [(x+(\Delta)x]^3 + \displaystyle [(x+(\Delta)x]^2 \displaystyle -$$\displaystyle (x^3+x^2)$ DIVIDED BY $\displaystyle (\Delta)x$

=$\displaystyle x^3+x^2(\Delta)x+x(\Delta)x^2+(\Delta)x^3+x^2+2x(\ Delta)x+(\Delta)x^2-x^3-x^2$ DIVIDED BY $\displaystyle (\Delta)x$

=$\displaystyle (\Delta)x[x^2+x(\Delta)x+(\Delta)x^2+2x+(\Delta)x]$ DIVIDED BY $\displaystyle (\Delta)x$

The $\displaystyle (\Delta)x$ will cancel leaving:

$\displaystyle [x^2+x(\Delta)x+(\Delta)x^2+2x+(\Delta)x]$

I plugged 0 in for $\displaystyle (\Delta)x$ and was left with

$\displaystyle x^2+2x$

But I know this is the wrong answer (the right answer should be $\displaystyle 3x^2+2x$)

**To clarify, DIVIDED BY $\displaystyle (\Delta)x$ means everything on that line is divided by $\displaystyle (\Delta)x$. I can't figure out how to write $\displaystyle (\Delta)x$ without the parentheses being around Delta. Nor can I figure out how to get $\displaystyle (\Delta)x$ in the denominator of the whole expression. I tried though.
You got the wrong answer because
$\displaystyle (a + b)^3 = (a + b) \cdot (a + b)^2$

$\displaystyle = (a + b) \cdot (a^2 + 2ab + b^2)$

$\displaystyle = a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3$

$\displaystyle = a^3 + 3a^2b + 3ab^2 + b^3$

-Dan

3. When I expand, I usually use Pascal's triangle. I know that the powers of the first term descend and second term ascend. For cubes, the coefficients are:
1 3 3 1

(x+y)^3 expands into:
$\displaystyle x^3 +x^2y + xy^2 + y^3$

Multiply by the coefficients:
$\displaystyle x^3 + 3x^2y + 3xy^2 + y^3$

Your mistake was that you didn't multiply by the coefficients. The binomial theorem is:
$\displaystyle (x+y)^n = \sum_{k=0}^{n}\ nCk\ x^{n-k}y^k$

You can use combinations to find the coefficients as well.

4. The thing that was throwing me off was cubing $\displaystyle [x+(\Delta)x]$. I know how to do $\displaystyle (a+b)^3$.

Would it be a big deal if I used another variable for $\displaystyle (\Delta)x$ and just plugged 0 in for that variable at the end?

Thanks for the responses and I figured out by trial and error that $\displaystyle [x+(\Delta)x]^3$ equals:

$\displaystyle x^3+3x^2(\Delta)x+3x(\Delta)x^2+(\Delta)x^3$

I guess I am seeing the two x's in the problem and getting confused.

5. If it helps you not get confused do it. Just sub the delta x back when you finish