How would you find the derivative of this function? I am doing some self studying and am trying to find the derivative of:
$$y= 2 \sqrt{x} $$ at the point [1, 2].
Now I know how to do the derivative the typical way, but could someone show me how to do the derivative using the $+h$ method?  I am having trouble solving the limit.  That is, limit of $h->0$ for:
$$\frac{2 \sqrt{1 + h} - 2  \sqrt{1}}{h}$$
All I get is infinity, which is clearly wrong as if you were to do derivatives the usual way, it would be:
$$\frac{dy}{dx} = \frac{1}{\sqrt{x}}$$
Equal to 1.
Thanks!
 A: I  suppose I should ask, what is the "usual way"?
$f'(x) = \lim_\limits{a\to x} \frac {2\sqrt {a} - 2\sqrt {x}}{a-x}$
We have substituted $x+h$ with $a.$ The algebra is nicer this way.
Multiply by $\frac {\sqrt {a} + \sqrt {x}}{\sqrt {a} + \sqrt {x}}$
$f'(x) = \lim_\limits{a\to x} \frac {2(a-x)}{(a-x)(\sqrt {a} + \sqrt {x})}$
The $a-x$ factors cancel when $a$ is close to but not equal to $x.$
$f'(x) = \lim_\limits{a\to x} \frac {2}{\sqrt {a} + \sqrt {x}}$
and let $a$ approach $x.$
$f'(x) = \frac {1}{\sqrt {x}}$
Alternatively, you can use the generalized binomial theorem.
$f'(x) = \lim_\limits{h\to 0} \frac {2(x+h)^\frac 12 - 2x^\frac 12}{h}$
The binomial theorem says:
$(x+h)^\frac 12 = x^\frac 12 + \frac 12 x^{-\frac 12}h - \frac 18 x^{-\frac 32}h^2 +\cdots$
The $n$th term is $c_n = c_{n-1} \frac {{\frac 12 - (n-1)}}{n} x^{-1}h$
$f'(x) = \lim_\limits{h\to 0} \frac {2x^\frac 12 + x^{-\frac 12}h - \frac 14 x^{-\frac 32}h^2 + \cdots - 2x^\frac 12}{h}\\
f'(x) = \lim_\limits{h\to 0} \frac {x^{-\frac 12} - \frac 14 x^{-\frac 32}h + \cdots}{h}$
And let $h$ go to $0.$
$f'(x) = \frac {1}{\sqrt {x}}$
A: Multiplying by the conjugate fraction $\frac{2\sqrt{1 + h} + 2}{2\sqrt{1 +h} + 2}$:
$$\lim_{h\rightarrow 0}\frac{2\sqrt{1 + h}-2}{h}\frac{2\sqrt{1+h}+2}{2\sqrt{1 + h} + 2} =\lim_{h\rightarrow 0}\frac{4h}{2h(\sqrt{1 + h} + 1)} = \lim_{h\rightarrow 0}\frac{2}{\sqrt{1+h} + 1}=\boxed{1}$$
When doing these derivatives, if you get a $\frac{0}{0}$ and you have roots in the numerator, it's usually best to multiply by the conjugate and simplify, as an $h$ will pop out somewhere.
