find derivatives "the long way" (using limits)

[R.W.]

My quiz on Monday requires us to do the long way of finding derivatives, using:

. . . . .f'(x) = lim_h->0 [f(x + h) - f(x)] / h

I used the limit for the equation y = x³ - 9. I know the derivative is supposed to be y' = 3x², according to the Power Rule. But when I plugged the values into the formula above, it didn't work. Instead, I got y' = x³ + 3x².

Here's my work:

. . . . .lim_h->0 [(x + h)³ - (x³ - 9)] / h

. . . . .lim_h->0 [x² + 2xh + h²(x + h) - (x³ - 9)] / h

. . . . .lim_h->0 [x³ + 3x²h + xh² + x²h + 2xh² + h³] / h

. . . . .lim_h->0 [x³ + 3x²h + 3xh² + h³] / h

The h on the bottom cancels out with the h in 3x²h, giving me:

. . . . .lim_h->0 x³ + 3x² + 3xh² + h³

Substitute 0 for the h's, and I get:

. . . . .x³ + 3x²

What did I do wrong?
_______________________
Edited by stapel -- Reason for edit: repairing formatting

 

[galactus]

\(\displaystyle \L\\\lim_{h\to\0}\frac{((x+h)^{3}-9)-(x^{3}-9)}{h}\)

\(\displaystyle \L\\\frac{3hx^{2}+3xh^{2}+h^{3}}{h}\)

\(\displaystyle \L\\\lim_{x\to\0}\frac{3x^{2}h}{h}+\lim_{x\to\0}\frac{3xh^{2}}{h}+\lim_{x\to\0}\frac{h^{3}}{h}\)

\(\displaystyle \H\\\lim_{x\to\0}{3x^{2}+3xh+h^{2}}=3x^{2}\)

 

[R.W.]

ok, thx i figured out that i didn't plug the whole eq. into f(x+h), i just did (x+h)^3, when i needed (x+h)^3 -9.

is there a way i can type like you? that looks a lot easier to read and make.

 

[tkhunny]

See "Forum Help" at the top of the screen. Read and practice the sections on "LaTeX".

 

[R.W.]

k thx i think i figured it out.