Need advice: what should be my next step for solving the derivative of $f(z)$ using the definition? What should be my next step for solving the derivative of $f(z)$ using the definition?
$$f(z) = (2{z^2} + 1) \cdot ({z^3} - \sqrt {z}) $$
$$f'(z) = \mathop {\lim }\limits_{\vartriangle z \to 0} \frac{{f(z + \vartriangle z) - f(z)}}{{\vartriangle z}} \\= \mathop {\lim }\limits_{\vartriangle z \to 0} \frac{{[2{{(z + \vartriangle z)}^2} + 1] \cdot [{{(z + \vartriangle z)}^3} - \sqrt {(z + \vartriangle z)} ] - (2{z^2} + 1) \cdot ({z^3} - \sqrt {z}) }}{{\vartriangle z}}$$
At this point I don't know how to proceed: when I try to expand it,  I can't isolate the ∆z to simplify and avoid the division by zero. What am I missing here?

After distributing the terms we have...
$$ f(z) = (2{z^2} + 1) \cdot ({z^3} - \sqrt {z}) = 2z^5 - 2z^{
\frac{5}{2}} + z^3 - z^{\frac{1}{2}} $$
by definition of derivative...
$$ f'(z) = \lim _{\Delta z \to 0} \frac{f(z+\Delta z) - f(z)}{\Delta z} $$
$$ = \lim _{\Delta z \to 0} \frac{2{(z+\Delta z)}^5 - 2(z+\Delta z)^{\frac{5}{2}} + (z+\Delta z)^3 - (z+\Delta z)^{\frac{1}{2}} - \Big[ 2z^5 - 2z^{
\frac{5}{2}} + z^3 - z^{\frac{1}{2}} \Big]}{\Delta z} $$
$$ = \lim _{\Delta z \to 0} \frac{2{(z+\Delta z)}^5 - 2(z+\Delta z)^{\frac{5}{2}} + (z+\Delta z)^3 - (z+\Delta z)^{\frac{1}{2}} - 2z^5 + 2z^{
\frac{5}{2}} - z^3 + z^{\frac{1}{2}}}{\Delta z} $$
You can proceed by factoring out the larger powers; solving for limits; and then factoring a $\Delta z$ from the remaining terms in the numerator. At that point, the $\Delta z$ in the numerator will cancel the one in the denominator, and you will be left with the answer. However, it is much easier to obtain the result using the standard differentiation rules, all of which can be proven using the definition of a limit. Otherwise, you are faced with the tedious task of computing an enormous polynomial.
For example, using standard differentation rules...
$$ f'(z) = \frac{d}{dz} f(z) = \frac{d}{dz} \Big[ 2z^5 - 2z^{
\frac{5}{2}} + z^3 - z^{\frac{1}{2}} \Big] $$
by the sum and difference rules we have
$$ = \frac{d}{dz} 2z^5 - \frac{d}{dz} 2z^{
\frac{5}{2}} + \frac{d}{dz} z^3 - \frac{d}{dz} z^{\frac{1}{2}} $$
by the constant multiple rule we have
$$ = 2 \frac{d}{dz} z^5 - 2 \frac{d}{dz} z^{
\frac{5}{2}} + \frac{d}{dz} z^3 - \frac{d}{dz} z^{\frac{1}{2}} $$
and by the power rule we have
$$ = 2 \cdot 5z^{(5-1)} - 2 \cdot  \frac{5}{2} z^{(
\frac{5}{2} - 1)} + 3 \cdot z^{(3-1)} - \frac{1}{2} \cdot z^{(\frac{1}{2}-1)} $$
and then we clean things up...
$$ = 10z^4 - 5z^{\frac{3}{2}} + 3z^2 - \frac{1}{2} z^{-\frac{1}{2}} $$
$$ = 10z^4 - 5z\sqrt{z} + 3z^2 - \frac{1}{2\sqrt{z}} $$
If you have any doubts about all of this, then I suggest reviewing the proofs of the standard differentiation rules. In particular, prove the sum/difference rule, the constant multiple rule, and power rule for any arbitrary function, using the definition of derivative. Then you will have confidence in this result without having to explicitly prove it from the ground up. You can find proof of said rules here.
