I'd like to find out if the function $f(x) = \sqrt{x^3}+3$ is differentiable for $\{x≥0 \ | \ x \in \mathbb{R}\}$ and if it is, then find it's derivative.
First I "simplified" the function $f(x)= (\sqrt {x^3}+3)^\frac{1}{2}$ and used the difference quotient to show that $f(x)$ is differentiable for $x>0$.
$$\frac{f(x) - f(x_0)}{x-x_0} = \frac{(\sqrt{x^3}+3)- \left(\sqrt{x_0^3}+3 \right)}{x-x_0}= \frac{\sqrt{x^3}- \sqrt{x_0^3}}{x-x_0}$$
We know that $x^3-x_0^3=\left(\sqrt{x^3} - \sqrt{x_0^3}\right)\left(\sqrt{x^3} + \sqrt{x_0^3}\right)$, therefore:
$$ \implies \frac{\left(\sqrt{x^3} - \sqrt{x_0^3} \right)}{\left({x^3} - \sqrt{x_0^3}\right)\left({x^3} + \sqrt{x_0^3}\right)} = \frac{1}{\left({x^3} + \sqrt{x_0^3}\right)}$$
Additionally, we show that $\lim_\limits{x\to x_0}\left({x^3} + \sqrt{x_0^3}\right) = 2\sqrt {x_0^3}$ and therefore:
$$\implies f'(x) = \frac{1}{2\sqrt {x_0^3}} $$
Therefore when $\{x>0 \ | \ x \in \mathbb{R}\}$, $f(x)$ is differentiable and now I'd have to show that $f(x)$ is differentiable for $x=0$. I do this by using the difference quotient.
$$\lim _\limits{x \to 0}\frac{f(x) - f(0)}{x-0}=\lim _\limits{x \to 0}\frac{\sqrt{x^3}+3 }{x}=\frac{\lim_\limits{x \to 0} \sqrt {x^3}+3}{\lim_\limits{x \to 0}x}$$
Since I get a situation where I would have to divide by $0$ I can't continue but in my exercise sheet it says that its $∞$ and I don't understand why. From the looks of it, I made a mistake somewhere or I have some understanding gap.
You start off fine, somewhere your algebra goes a little wonky.
You are not quite explict with that algebra. So, I am having a had time putting my finger on the error.
This is how I would do it.
$f'(a) = \lim_\limits{x\to a} \frac {f(x) - f(a)}{x-a}\\ \lim_\limits{x\to a} \frac {(\sqrt{x^3}+3) - (\sqrt{a^3}+3)}{x-a}\\ \lim_\limits{x\to a} \frac {\sqrt{x^3} - \sqrt{a^3}}{x-a}\\ \lim_\limits{x\to a} \frac {(\sqrt{x^3} - \sqrt{a^3})(\sqrt{x^3} + \sqrt{a^3})}{(x-a)(\sqrt{x^3} + \sqrt{a^3})}\\ \lim_\limits{x\to a} \frac {x^3 - a^3}{(x-a)(\sqrt{x^3} + \sqrt{a^3})}\\ \lim_\limits{x\to a} \frac {(x-a)(x^2+ax+a^2)}{(x-a)(\sqrt{x^3} + \sqrt{a^3})}\\ f'(a) =\frac {3a^2}{2a\sqrt{a}}\\ f'(a) =\frac {3\sqrt a}{2}\\ f'(0) =0\\ $