# Find derivative using limit definition

• Feb 28th 2013, 07:35 AM
Ragnarok
Find derivative using limit definition
Find the derivative of $\displaystyle f(z)=z^4-z^2$ using the following definition:

$\displaystyle f'(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}$

I tried

$\displaystyle f'(z_0)=\lim_{z\to z_0}\frac{(z^4-z^2)-(z_0^4-z_0^2)}{z-z_0}$

$\displaystyle =\lim_{z\to z_0}\frac{z^2(z+1)(z-1)-z_0^2(z_0+1)(z_0-1)}{z-z_0}$

but am unclear how to proceed.
• Feb 28th 2013, 07:55 AM
MINOANMAN
Re: Find derivative using limit definition
Ragnarok

Substitute z with z=z0+δz and do all the necessary expansions and simplifications .then the denominator will become δz .
factorize the numerator and cancel δz . then the limit will show you the final result that must be 4z0^3-2z0
try it
• Feb 28th 2013, 09:33 AM
Ragnarok
Re: Find derivative using limit definition
Thanks Minoanman, but I was specifically instructed to use the above "alternative" definition of derivative as opposed to the one using $\displaystyle f(z_0+\Delta z)-f(z_0)$. Could anyone help me with the algebra/tell me what I'm doing wrong?
• Feb 28th 2013, 09:59 AM
Plato
Re: Find derivative using limit definition
Quote:

Originally Posted by Ragnarok
Find the derivative of $\displaystyle f(z)=z^4-z^2$ using the following definition:

$\displaystyle f'(z_0)=\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}$

I tried

$\displaystyle f'(z_0)=\lim_{z\to z_0}\frac{(z^4-z^2)-(z_0^4-z_0^2)}{z-z_0}$

$\displaystyle =\lim_{z\to z_0}\frac{z^2(z+1)(z-1)-z_0^2(z_0+1)(z_0-1)}{z-z_0}$

but am unclear how to proceed.

$\displaystyle \frac{(z^4-z^2)-(z_0^4-z_0^2)}{z-z_0}=\frac{(z^4-z_0^4)-(z^2-z_0^2)}{z-z_0}$

Now factor. The idea is divide off the $\displaystyle z-z_0$ factor.
• Feb 28th 2013, 10:10 AM
Ragnarok
Re: Find derivative using limit definition
Brilliant, thank you!
• Feb 28th 2013, 12:10 PM
Soroban
Re: Find derivative using limit definition
Hello, Ragnarok!

You started correctly . . .

Quote:

$\displaystyle \text{Find the derivative of }f(x)\,=\,x^4-x^2\,\text{ at }x=a$

$\displaystyle \text{ using this definition: }\:f'(a)\:=\:\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$

I like to break it up into three steps.

$\displaystyle [1]\;f(x) - f(a) \;=\; (x^4 - x^2) - (a^4-a^2)$

. . . . . . . . . . . $\displaystyle =\; (x^4-a^4) - (x^2 - a^2)$

. . . . . . . . . . . $\displaystyle =\;(x-a)(x+a)(x^2+a^2) - (x-a)(x+a)$

. . . . . . . . . . . $\displaystyle =\;(x-a)\,\big[(x+a)(x^2+a^2) - (x+a)\big]$

$\displaystyle [2]\;\frac{f(x)-f(a)}{x-a} \;=\;\frac{(x-a)\big[(x+a)(x^2+a^2) - (x+a)\big]}{x-a}$

. . . . . . . . . . . $\displaystyle =\;(x+a)(x^2+a^2) - (x+a)$

$\displaystyle [3]\;\lim_{x\to a}\frac{f(x)-f(a)}{x-a} \;=\;\lim_{x\to a}\big[(x+a)(x^2+a^2) - (x+a)\big]$

. . . . . . . . . . . . . . $\displaystyle =\; (a+a)(a^2+a^2) - (a+a)$

. . . . . . . . . . . . . . $\displaystyle =\;(2a)(2a^2) - 2a$

Therefore: . .$\displaystyle f'(a) \;=\;4a^3 - 2a$