# How do you find the slope of the tangent line to the graph of the equation at the point with x-coordinate for y = sqrt(x) at p(4,2)?

Jun 10, 2015

You first find the derivative and then use that at ${x}_{p} = 4$

#### Explanation:

$y = \sqrt{x} = {x}^{\frac{1}{2}} \to$

$y ' = \frac{1}{2} {x}^{- \frac{1}{2}} = \frac{1}{2 \cdot \sqrt{x}}$

Using ${x}_{p} = 4$:

$y ' \left(4\right) = \frac{1}{2. \sqrt{4}} = \frac{1}{4}$

And this is the slope.