Find $f'(x)$ at given value of x Find $f'(x)$ at the given value of x
$f(x)=\sqrt{x+2}$ 
Find $f'(7)$
My question for this one is do I approach this question by trying to find the derivative of the initial equation and then once I have found the derivative do I simply plug in 7 for x and solve?
So once I start solving I would get $\frac{\sqrt{x+h+2}-\sqrt{x+2}}{h}$
$\lim_{h\to0} \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h} = \lim_{h\to0} \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h} \cdot \frac{\sqrt{x+h+2}+\sqrt{x+2}}{\sqrt{x+h+2}+\sqrt{x+2}}$
From that point I would get 
$\frac{(x+h+2-x-2)}{h\sqrt{x+h+2}+\sqrt{x+2}}$
 A: Yes. That is exactly how you do it!
Find $f'(x)$, then plug in $x=7$ into the resulting function and compute. 
Here's how:
$$
\begin{align*}
f'(x)&=\lim_{h\to0} \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h} \\
&= \lim_{h\to0} \frac{(\sqrt{x+h+2}-\sqrt{x+2})}{h} \cdot \frac{(\sqrt{x+h+2}+\sqrt{x+2})}{(\sqrt{x+h+2}+\sqrt{x+2})}\\
&=\lim_{h\to0}\frac{x+h+2-x-2}{h(\sqrt{x+h+2}+\sqrt{x+2})}\\
&=\lim_{h\to 0} \frac{1}{\sqrt{x+h+2}+\sqrt{x+2}}\\
f'(x)&=\frac{1}{2\sqrt{x+2}}
\end{align*}
$$
Now plugging in $x=7$ gives: $$f'(7)=\frac{1}{2\sqrt{7+2}}=\frac{1}{6}.$$
A: $$f'(x)=\frac{1}{2\sqrt{x+2}}$$
$$f'(7)=\frac{1}{2\sqrt{7+2}}=\frac{1}{6}$$
A: Hint: Since you're using the difference quotient, you want to get rid of the square roots in the numerator by using the numerator's conjugate:
$$\lim_{h\to0} \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h} = \lim_{h\to0} \frac{\sqrt{x+h+2}-\sqrt{x+2}}{h} \cdot \frac{\sqrt{x+h+2}+\sqrt{x+2}}{\sqrt{x+h+2}+\sqrt{x+2}}$$
A: The way you sketch -- first find $f'(x)$ in general -- would be right if you had a rule that allowed you to do that easier than going to the definition.
Specifically, for $f(x)=\sqrt{x+2}$ you would use the chain rule and the power rule that the deriviative of $x^n$ is $nx^{n-1}$, here with $n=\frac12$.
But if you don't have these two rules available, then there's no real reason to keep $x$ a variable while finding the limit. That's just more work that won't benefit you in this particular context. So since
$$ f'(x) = \lim_{h\to 0} \frac{\sqrt{x+2+h}-\sqrt{x+2}}h $$
the first thing you should do, before worrying about the limit, is to plug in $x=7$:
$$ f'(7) = \lim_{h\to 0} \frac{\sqrt{9+h}-\sqrt{9}}h $$
and then you can look for the limit without being unduly distracted by the $x$.
(Hint for that: multiply by $\frac{\sqrt{9+h}+\sqrt 9}{\sqrt{9+h}+\sqrt{9}}$).
A: Note that the numerator is the difference of two square roots. You can simplify this expression by multiplying the numerator and the denominator by the sum of the square roots. 
