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. (Happy)