Finding a derivative.

**JennyFlowers** · March 21st 2011, 12:33 PM

I'm not finding correctly the derivative of:

$f(x) = x^4 - 1$

I do the following:

lim h-> 0:

$\frac{(x+h)^4 -1 -(x^4-1)}{h}$

$= \frac{(x^2+h^2)^2 -x^4}{h}$

$= \frac{x^4 + 2x^2h^2 + h^4 - x^4}{h}$

$= \frac{2x^2h^2+h^4}{h}$

$= 2x^2h+h^3$

Then, of course, when I substitute "0" for "h", I get an answer of 0.

The correct answer is: $4x^3$

Where did I go wrong?

Thank you!

**pickslides** · March 21st 2011, 12:50 PM

In the second line of your workings...

$\displaystyle (x+h)^4 = (x+h)(x+h)(x+h)(x+h) = x^4+4x^3h+\dots +h^4 \neq (x^2+h^2)^2$

**JennyFlowers** · March 21st 2011, 01:08 PM

Is there a simpler way to solve this while still using the definition of the derivative? I'm sure I did this before without expanding it the way you have, I just don't remember how.

Thanks again.

**JennyFlowers** · March 21st 2011, 01:22 PM

Ah, found it. I used the alternate definition of the derivative to do it.

Works out to:

$\frac{x^4-c^4}{x-c}$

$\frac{(x+c)(x-c)(x^2+c^2)}{x-c}$

$(x+c)(x^2+c^2)$

Which solves my problem as well as the derivative from the standard definition.

Thanks for the help!

**pickslides** · March 21st 2011, 01:33 PM

Originally Posted by JennyFlowers

Is there a simpler way to solve this while still using the definition of the derivative?

$\displaystyle (x^{n})'= nx^{n-1}$

**JennyFlowers** · March 21st 2011, 01:37 PM

Right, but I wasn't allowed to find this derivative using the power rule. I was required to find it using either the definition or the alternate definition!

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