Finding the equation of tangent line to the curve at the given point 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 $(\sqrt{x} - 1)/1-1$
 A: 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).
A: 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 $x$.

A: 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
