# Cubing (x+∆x)

• September 12th 2008, 05:54 PM
yeloc
Cubing (x+∆x)
We just started differential equations in my AP Calculus class, and the only thing I am having trouble with is $[(x+(\Delta)x]^3$. 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 $(x+(\Delta)x)$.

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

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

My work thus far:

f' (x)= $\lim_{(\Delta)x\rightarrow0}$ $[(x+(\Delta)x]^3$ + $[(x+(\Delta)x]^2$ $-$ $(x^3+x^2)$ DIVIDED BY $(\Delta)x$

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

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

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

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

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

$x^2+2x$

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

**To clarify, DIVIDED BY $(\Delta)x$ means everything on that line is divided by $(\Delta)x$. I can't figure out how to write $(\Delta)x$ without the parentheses being around Delta. Nor can I figure out how to get $(\Delta)x$ in the denominator of the whole expression. I tried though. (Happy)
• September 12th 2008, 06:00 PM
topsquark
Quote:

Originally Posted by yeloc
We just started differential equations in my AP Calculus class, and the only thing I am having trouble with is $[(x+(\Delta)x]^3$. 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 $(x+(\Delta)x)$.

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

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

My work thus far:

f' (x)= $\lim_{(\Delta)x\rightarrow0}$ $[(x+(\Delta)x]^3$ + $[(x+(\Delta)x]^2$ $-$ $(x^3+x^2)$ DIVIDED BY $(\Delta)x$

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

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

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

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

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

$x^2+2x$

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

**To clarify, DIVIDED BY $(\Delta)x$ means everything on that line is divided by $(\Delta)x$. I can't figure out how to write $(\Delta)x$ without the parentheses being around Delta. Nor can I figure out how to get $(\Delta)x$ in the denominator of the whole expression. I tried though. (Happy)

You got the wrong answer because
$(a + b)^3 = (a + b) \cdot (a + b)^2$

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

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

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

-Dan
• September 12th 2008, 06:00 PM
Chop Suey
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:
$x^3 +x^2y + xy^2 + y^3$

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

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

You can use combinations to find the coefficients as well.
• September 12th 2008, 06:33 PM
yeloc
The thing that was throwing me off was cubing $[x+(\Delta)x]$. I know how to do $(a+b)^3$.

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

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

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

I guess I am seeing the two x's in the problem and getting confused.(Thinking)
• September 12th 2008, 06:39 PM
11rdc11
If it helps you not get confused do it. Just sub the delta x back when you finish