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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.

egreg
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Rahul
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  • The equation of the tangent is $y-f(3)=f'(3)\cdot(x-3)$, by definition of tangent. – egreg Dec 22 '15 at 22:07
  • The tangent is the straight line passing through the point of tangency $(3,f(3))$ and whose slope is the same as the function at that point: $f'(3)$ – MasB Dec 22 '15 at 22:14

1 Answers1

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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 :)

T. Porter
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  • still don't know how to make graph... – Rahul Dec 22 '15 at 22:51
  • The main job for plotting a line is basically the equation. If you know the equation, you can proceed by a fixed scheme. The slope intercept form is a very common form, so there are mass of good guides for this.

    For example look at:

    https://www.mathsisfun.com/equation_of_line.html or http://www.webmath.com/gline.html

     – T. Porter Dec 22 '15 at 23:02
  • https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-linear-equations-functions/8th-slope/v/graphing-a-line-in-slope-intercept-form shows it in a video. It uses exaktly the slope intercept form, except that they print y instead of f(x) and b instead of t – T. Porter Dec 22 '15 at 23:06