Derivative of a complex function using limit definition

**Ragnarok** · February 28th 2013, 06:10 AM

Find the derivative of $f(z)=5z^2-10z+8$ using the following definition:

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

I started with

$f'(z_0)=\lim_{z\to z_0}\frac{5z^2-10z+8-(5z_0^2-10z_0+8)}{z-z_0}$

which simplifies a bit to

$f'(z_0)=\lim_{z\to z_0}\frac{5z^2-10z-5z_0^2+10z_0}{z-z_0}$

but where do you go from there? Thanks!

**Plato** · February 28th 2013, 06:38 AM

Originally Posted by Ragnarok

Find the derivative of $f(z)=5z^2-10z+8$ using the following definition:

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

I started with

$f'(z_0)=\lim_{z\to z_0}\frac{5z^2-10z+8-(5z_0^2-10z_0+8)}{z-z_0}$

which simplifies a bit to

$f'(z_0)=\lim_{z\to z_0}\frac{5z^2-10z-5z_0^2+10z_0}{z-z_0}$

but where do you go from there? Thanks!

$\frac{5z^2-10z-5z_0^2+10z_0}{z-z_0}=\frac{5(z^2-z_0^2)-10(z-z_0)}{z-z_0}$

**hollywood** · February 28th 2013, 06:53 AM

Try this:

$f'(z_0)=\lim_{z\to z_0}\frac{5z^2-5z_0^2-10z+10z_0}{z-z_0}$

$f'(z_0)=\lim_{z\to z_0}\frac{5(z+z_0)(z-z_0)-10(z-z_0)}{z-z_0}$

- Hollywood

Awww - Plato, you beat me!

**Ragnarok** · February 28th 2013, 06:55 AM

Ah thank you! I forgot my difference of squares factorization.

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