# Thread: find derivatives "the long way" (using limits)

1. ## find derivatives "the long way" (using limits)

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

. . . . .f'(x) = lim<sub>h->0</sub> [f(x + h) - f(x)] / h

I used the limit for the equation y = x<sup>3</sup> - 9. I know the derivative is supposed to be y' = 3x<sup>2</sup>, according to the Power Rule. But when I plugged the values into the formula above, it didn't work. Instead, I got y' = x<sup>3</sup> + 3x<sup>2</sup>.

Here's my work:

. . . . .lim<sub>h->0</sub> [(x + h)<sup>3</sup> - (x<sup>3</sup> - 9)] / h

. . . . .lim<sub>h->0</sub> [x<sup>2</sup> + 2xh + h<sup>2</sup>(x + h) - (x<sup>3</sup> - 9)] / h

. . . . .lim<sub>h->0</sub> [x<sup>3</sup> + 3x<sup>2</sup>h + xh<sup>2</sup> + x<sup>2</sup>h + 2xh<sup>2</sup> + h<sup>3</sup>] / h

. . . . .lim<sub>h->0</sub> [x<sup>3</sup> + 3x<sup>2</sup>h + 3xh<sup>2</sup> + h<sup>3</sup>] / h

The h on the bottom cancels out with the h in 3x<sup>2</sup>h, giving me:

. . . . .lim<sub>h->0</sub> x<sup>3</sup> + 3x<sup>2</sup> + 3xh<sup>2</sup> + h<sup>3</sup>

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

. . . . .x<sup>3</sup> + 3x<sup>2</sup>

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

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

$\L\\\frac{3hx^{2}+3xh^{2}+h^{3}}{h}$

$\L\\\lim_{x\to\0}\frac{3x^{2}h}{h}+\lim_{x\to\0}\f rac{3xh^{2}}{h}+\lim_{x\to\0}\frac{h^{3}}{h}$

$\H\\\lim_{x\to\0}{3x^{2}+3xh+h^{2}}=3x^{2}$

3. 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.

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

5. k thx i think i figured it out.

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