How do I calculate the derivativee of the function $x\mapsto \sqrt{x}$ in two ways? I need to calculate the derivative of the following function in two ways:
$$
f\colon\mathbb{R}^+\to\mathbb{R}^+,
\quad x\mapsto \sqrt{x}
$$
a) by means of differential quotient
b) using the derivation rules for powers
I need to determine the equation of the tangent of the graph of $f$ at the point $(3,f(3))$. Sketch the graph of $f$ and the tangent. Can someone do this?
a) I know that and I get $1/(\sqrt{x}+\sqrt{x})$
b) $1/(2\sqrt{x})$
But what they mean with “determine the equation of the tangent”? How to do this last part? I would be really thankful if someone could help me.
 A: Regarding the tangent:
The tangent of the graph of $f$ at the point $P=(3, f(3))$ is the line with the same slope as the graph has in $P$ and going through the point $P$.
Therefore starting with the general line equation $f_l(x) = m x + t$, $m$ denotes the slope and $t$ the y-intercept.
To achieve the same slope as the graph of $f$ in $P$ just set $m = f'(P_x)$, so to the value of the derivation in the specified point (to $m = 1/(2 \sqrt{P_x})$ with $P_x = 3$ in your case, so $m = 1/\sqrt{6}$).
To get $t$ one can use the general line equation again $y = f_l(x) = mx + t$, which is true for every point, so also for $P$, which leads us to $P_y = m P_x + t$, or $t = P_y - m P_x$, with $P_x = 3$ and $P_y = f(3) = \sqrt{3}$ in your case, so $$t = \sqrt{3} - (1/\sqrt{6}) 3 = \sqrt{6}/\sqrt{2} - (\sqrt{3}\sqrt{3})/\sqrt{6} = \sqrt{6}/\sqrt{2} - \sqrt{3}/\sqrt{2} = $$
$$ = (\sqrt{6} - \sqrt{3})/\sqrt{2} = (\sqrt{2} - 1) \sqrt{3/2}$$
(however you like to express it)
This results in the tangent of the graph of $f$ in $P$: $f_t(x) = x/\sqrt{6} + (\sqrt{6} - \sqrt{3})/\sqrt{2}$
Regarding the plot:
Every line with an equation given in the form $f(x) = m x + t$ can be plot simply by starting at the point $(0|t)$ (Value t on the y axis) and giving the line a slope of $m$ ($n$ units in direction of the x axis leading to $n m$ units in the direction of the y axis).
Preferable here:
As you know the tangent goes through the point $P$, you can also just start off the point $P$ and apply a slope of $m$ again.
Hope that helped you :)
