I have this piece-wise function:
$$f(x)=\begin{cases} 2x+3 & \text{x$\le4$} \\ x^2-5 &\text{x$\gt 4$}\end{cases}$$
I know the function is continuous at $x=4$ except how would I go finding out if it were differentable at that point?
I tried by taking the derivative of the top portion and the bottom portion and got
$$2$$ $$2x$$
Then I plugged in $4$ for both and got $2$, and $8$. Since $2 \ne 8$ I deduced that at $x=4$ the function is not differentable. Is my ideology correct of am I wrong?
What you can immediately deduce from your observation is that the function cannot be continuously differentiable (in that point). It's also not difficult to see that it also cannot be differentiable, but strictly speaking (esp if you are not sure about what you just wrote down) you would need to/should either check the definition or make use of some theorem which contradicts your observation.
You can, e.g., just use the fact that you would need to have $\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} $ has to exist and to be $=f^\prime(4)$ regardless of whether $h$ approaches $4$ from left or right. In the first case the limit is $2$, in the second case it will be $8$.