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 $(\sqrt{x} - 1)/1-1$
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 $(\sqrt{x} - 1)/1-1$
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).
Several comments:
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