help with derivative

**georgel** · Aug 26th 2009, 09:03 AM

I need help with the following expression:

$d/dz (z^2+3i)$

Thank you.

**eXist** · Aug 26th 2009, 09:19 AM

Well just break it up:

$d/dz(z^2) + d/dz(3i)$
$= 2z + 0$
$= 2z$

In this case 3i is a constant, so the derivative is 0.

**georgel** · Aug 26th 2009, 09:28 AM

Thank you!!!
What if it were a bit more complicated?
For example,

$\frac{(z-3i)^2}{(z^2+4)^2}$ ?

**arbolis** · Aug 26th 2009, 09:42 AM

Originally Posted by georgel

Thank you!!!
What if it were a bit more complicated?
For example,

$\frac{(z-3i)^2}{(z^2+4)^2}$ ?

I do not really know.
I know you can always use the definition, i.e. $f'(z)= \lim _{h \to 0} \frac{f(z+h)-f(z)}{h}$ where $f(z)=\frac{(z-3i)^2}{(z^2+4)^2}$ . Of course $h\in \mathbb{C}$ .

**eXist** · Aug 26th 2009, 09:55 AM

$\frac{(z-3i)^2}{(z^2+4)^2}$

If it were something like this, then you can rewrite it as so:

$(z-3i)^2(z^2+4)^{-2}$

From here you apply the product rule. (You'll have a chain rule to apply as well.)

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