Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am trying to find the tangent line at $y=\sqrt{x} $ , (1,1)

I know that I need to use the tangent line equation and I end up with $(\sqrt{x} - 1)/1-1$

share|improve this question
Are you trying to find the tangent line or the slope of the tangent line? There is a big difference. The tangent line has an equation ($y = mx + b$ or something), and the slope of the tangent line is a number. –  Jesse Madnick Sep 11 '11 at 17:52
Note that the equation of a line will have shape $ax+by=c$, where $a$, $b$, $c$ are constants. Anything that cannot be put into that shape cannot be right. –  André Nicolas Sep 11 '11 at 17:57
"Find an equation of the tangent line to the curve at the given point." Not sure what the differences are in what you mentioned. –  user138246 Sep 11 '11 at 17:58
@Jordan: I believe André is saying that there is a difference between "an equation of the tangent line to a curve" and the number given by $\lim_{x\to 1}\frac{\sqrt{x} - 1}{x - 1}.$ –  Jesse Madnick Sep 11 '11 at 22:46
add comment

3 Answers

up vote 5 down vote accepted

EDIT: Let's clarify a couple of things.

The slope of the secant line between $(a, f(a))$ and $(x,f(x)))$ is $$\frac{f(x) - f(a)}{x-a}.$$

The slope of the tangent line at $(a, f(a))$ is $$\lim_{x\to a}\frac{f(x) - f(a)}{x-a}.$$

To find the equation of a tangent line, one needs to use the point-slope formula, which I've explained below.

Now, in your case, $f(x) = \sqrt{x}$, and we have $a = 1$, $f(a) = 1$. So the slope of the tangent line is $$\lim_{x\to 1}\frac{\sqrt{x} - 1}{x-1}.$$ Now we have to evaluate this limit.

If we try to evaluate this limit by just plugging in $x = 1$, we get $0/0$, which is a problem (dividing by zero is bad), so we need a new strategy.

Idea: When evaluating the limits of fractions, a good trick is to multiply the top and bottom by the "radical conjugate." So:

$$\begin{align} \frac{\sqrt{x} - 1}{x-1} & = \frac{\sqrt{x} - 1}{x-1}\frac{\sqrt{x} + 1}{\sqrt{x} + 1} \\ & = \frac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{(x-1)(\sqrt{x} + 1)} \\ & = \frac{x - 1}{(x-1)(\sqrt{x} + 1)} \\ & = \frac{1}{\sqrt{x} + 1}. \end{align}$$

Now we can evaluate $$\lim_{x \to 1} \frac{\sqrt{x} - 1}{x-1} = \lim_{x\to 1}\frac{1}{\sqrt{x} + 1}$$ by plugging in $x = 1$ no problem. This will give us the slope of the tangent line. If you want the equation of the tangent line, you need the point-slope formula, explained below.

The point-slope formula says that a line with slope $m$ that passes through $(x_0, y_0)$ has an equation of the form $$y - y_0 = m(x-x_0).$$

In your case, the tangent line passes through $(1,1)$, so you can plug in $x_0 = 1$, $y_0 = 1$. We'll also have the slope, $m$, from the previous section once we evaluate that limit (which I leave to you to do).

share|improve this answer
Well in my book is explains to me to by using the equation (f(x)-f(a))/(x-a) I tried to do that and I get something that doesn't make sense. –  user138246 Sep 11 '11 at 17:46
@Jordan: Note that basically $y = f(x)$ and $y_0 = f(x_0)$, so Jesse's answer says $m = (f(x) - f(x_0))/(x - x_0)$. Replacing $x_0$ with $a$ gives the same expression as your book. –  TMM Sep 11 '11 at 17:49
I see. But are you sure that the formula $(f(x) - f(a))/(x-a)$ really gives you a formula for the tangent line? Because really, $(f(x) - f(a))/(x-a)$ gives the slope of the secant line between $(a, f(a))$ and $(x,f(x))$. On the other hand, the slope of the tangent line is $$\lim_{x\to a} \frac{f(x) - f(a)}{x-a}.$$ Note that neither of these formulas give an equation for a line, but rather a formula for computing the slope of a line. –  Jesse Madnick Sep 11 '11 at 17:49
Yes it does, does that mean I make a = x? –  user138246 Sep 11 '11 at 17:50
Neither Jesse's nor Austin's comments were rude or offensive. Do not flag them as such. –  Zev Chonoles Sep 11 '11 at 18:30
show 10 more comments

Several comments:

  • You shouldn't write $(\sqrt{x} - 1)/1-1$ if you mean $(\sqrt{x} - 1)/(1-1)$; remember the conventions on order of operations.
  • If you put $1$ in place of $x$ in $(\sqrt{x} - 1)/(x-1)$, what you get is $(\sqrt{1} - 1)/(1-1)$. This is $0/0$.
  • The expression $(\sqrt{x} - 1)/(x-1)$ gives the slope of a secant line, not of a tangent line.
  • Since $0/0$ is undefined, in order to find the slope of the tangent line, you need to find $\lim\limits_{x\to1} (\sqrt{x} - 1)/(x-1)$, rather than simply plugging in $1$ in place of $x$.
share|improve this answer
My biggest problem is I don't know what f(x) is and what f(a) is, is is y and x or what? My book does not explain this anywhere or even show work for anything really, they just jump to the answer. Every example the book uses a = 1 but the points are always 1,1. Whoever made this book is an idiot, there is no definition of 1 in this book. Is is either x or y. –  user138246 Sep 11 '11 at 18:19
add comment

y=root of x, or simply f(x)= root of x at the point (1,1) by deriving the root of x we get x^1/2 which is equal to 1/2.root of x plug in the x value from (1,1) and get 1/2 we know y,m,and x so we only need to find c in this equation, y=mx+c then we get 1=1/2*1+c c=1/2 so our equation will be y=1/2x+1/2

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.