Find f'(x) at the given value of x Find $f^\prime (x)$ at the given value of $x$:
$f(x)=3\sqrt{x}$
Find $f^\prime (5)$
For this one in my attempt to find the derivative I ended up with "$9/0$" which would lead me to believe that the value does not exist at $f^\prime (5)$. I was wondering if this was correct and if not how I would approach this question differently.
$\lim_{h\to0} \frac{3\sqrt{x+h}-3\sqrt{x}}{h} \cdot \frac{3\sqrt{x+h}-3\sqrt{x}}{3\sqrt{x+h}-3\sqrt{x}}$
$\lim_{h\to0} \frac{9(x+h)-9x}{h(3\sqrt{x+h}{-3\sqrt{x}})}$

Since you're using the formal definition of the derivative here, given $f(x)=3\sqrt{x}$, we know that
$$
\begin{align*}
f'(x)&=\lim_{h\to0} \frac{f(x+h)-f(x)}{h} \\
&= \lim_{h\to0} \frac{3\sqrt{x+h}-3\sqrt{x}}{h}\\
&= 3 \cdot \lim_{h\to0} \frac{\sqrt{x+h}-\sqrt{x}}{h}\\
\end{align*}
$$
We can then multiply the numerator and the denominator by the conjugate:
$$
f'(x)= 3 \cdot \lim_{h\to0} \frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\\
$$
Try simplifying from here.
