I am suppose to find the equation of a tangent line and I am given the following information: $ y = \sqrt[4] {x}$, $(1,1)$ I know the formula $f(a+h)-f(a)$ but it does not seem to be helping me.

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    $\begingroup$ The equation of the tangent line to the graph of $f(x)$ at $(a,f(a))$ is $$y=f(a)+f'(a)(x-a),$$ where $f'(a)$ is the derivative of $f(x)$ at $x=a$. By definition this derivative is $$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.$$ In your case $f(x)=\sqrt[4]{x}$ and $a=1$. $\endgroup$ – Américo Tavares Sep 18 '11 at 0:56
  • $\begingroup$ Shouldn't x-a be zero? $\endgroup$ – toby yeats Sep 18 '11 at 1:16
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    $\begingroup$ $x-a=h$ tends to $0$. $\endgroup$ – Américo Tavares Sep 18 '11 at 1:36
  • $\begingroup$ $$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$ $\endgroup$ – Américo Tavares Sep 18 '11 at 1:40
  • $\begingroup$ I am a little confused as to what I need to do because it is actually not even in this book that I can see. So I need to first find the derivative, then plug in the x and then add that into the slope for the point slope form? $\endgroup$ – toby yeats Sep 18 '11 at 1:45

Equation of tangent line at point $(a,f(a))$ is $y = f(a) + f'(a)(x - a)$, so we have to find $f'(x)$ and than plug in value $a$ into the result.

$$ f'(x) = (\sqrt[4]{x})' = ({x^{\frac{1}{4}}})' = \frac{1}{4}{x^{\frac{{ - 3}}{4}}}$$

$$\implies f'(a) = f'(1) = \frac{1}{4}$$

Since $f(1) = 1$, we can write next equation $y = 1 + \frac{1}{4}(x - 1)$

which means that equation of tangent line is $y = \frac{1}{4}x + \frac{3}{4}$.

  • $\begingroup$ Why this impulse you have of posting the entire answer as a $\LaTeX$ equation in sundry ways? It also seems to screw up the search: I tried searching for "Equotion" (the misspelling you have twice), but search could not find it. By contrast, it was able to find the response by mixedmath when searching for "SET OF HINTS". So, could you please stop it? $\endgroup$ – Arturo Magidin Sep 18 '11 at 4:01
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    $\begingroup$ And now that I have edited it, search finds the reply when I search for "Equotion". I think that settles it: it is activiely detrimental to the site for you to post your entire answer, text and all, as a $\LaTeX$ equation using \text; it makes your answers unsearchable. $\endgroup$ – Arturo Magidin Sep 18 '11 at 4:05
  • $\begingroup$ pedja, I assumed that you somehow misunderstood that the entire equation should be posted as a $\LaTeX$ equation. So I edited it. (Later I found that @Arturo had also edited it before me.) $\endgroup$ – Srivatsan Sep 18 '11 at 4:07
  • $\begingroup$ @ArturoMagidin,I am using MathType editor so that is probably reason of such LATex conversion.If you think that such formating cause problems I will stop $\endgroup$ – Peđa Terzić Sep 18 '11 at 4:09
  • $\begingroup$ @pedja: You are posting your text as LaTeX equation, using \text. (To make it worse, you also using it haphazardly on variable names, so that sometimes you typeset $\text{m}$ and sometimes $m$, sometimes $x$ and sometimes $\text{x}$. Yes, it causes problems, and having the entire post be a bunch of $\LaTeX$ processed by MathJax makes it "invisible" to the search engine. It means nobody can find the post by searching for content. So, yes, please stop. $\endgroup$ – Arturo Magidin Sep 18 '11 at 4:16


For the point-slope form of a tangent lint, $y - y_0 = m (x - x_0)$, you need exactly 2 pieces of information: the slope $m$ of the line and a point $(x_0, y_0)$ that the line goes through.

You have the point, so how do you find the slope? Well, the derivative tells you something about the slope, right? If you know what that is, then you can assemble both of these pieces of information into the equation for the tangent line.

  • $\begingroup$ @Jordan: every equation for a line that I know has an x and a y in it (this is understatement- they all have variables). No? $\endgroup$ – davidlowryduda Sep 18 '11 at 2:03
  • $\begingroup$ y-y1=m(x-x1) I am just not sure how to use it here. $\endgroup$ – toby yeats Sep 18 '11 at 2:12
  • $\begingroup$ @Jordan: Well, then I suppose you should look into it more. $\endgroup$ – davidlowryduda Sep 18 '11 at 2:15
  • $\begingroup$ I mean I know that the equation is y-1=deriviative (x-1) but I can't get the right answer from that. $\endgroup$ – toby yeats Sep 18 '11 at 2:16
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    $\begingroup$ I have the correct answer in the back of my book, but not the ability to get to it on my own yet. $\endgroup$ – toby yeats Sep 18 '11 at 2:24

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