# How do you find the equation of the tangent and normal line to the curve y=x^(1/2) at x=1?

Jun 10, 2018

#### Explanation:

We find the tangent line to a curve by finding the derivative first.

$\frac{d}{\mathrm{dx}} \left({x}^{n}\right) = n {x}^{n - 1}$ given that $n$ is a constant.

$f \left(x\right) = {x}^{1} / 2$

$\implies f ' \left(x\right) = \frac{1}{2} \cdot {x}^{1 - \frac{1}{2}}$

$\implies f ' \left(x\right) = \frac{1}{2} \cdot {x}^{- \frac{1}{2}}$

$\implies f ' \left(x\right) = \frac{1}{2} \cdot \frac{1}{\sqrt{x}}$

$\implies f ' \left(x\right) = \frac{1}{2 \sqrt{x}}$

$\implies f ' \left(1\right) = \frac{1}{2 \sqrt{1}}$

$\implies f ' \left(1\right) = \frac{1}{2}$

Let's now find the y value when $x = 1$

$\implies f \left(1\right) = \sqrt{1}$

$\implies f \left(1\right) = 1$

We can now use these information to find the equation of the tangent line using the formula $m \left(x - {x}_{1}\right) = y - {y}_{1}$

$\implies \frac{1}{2} \left(x - 1\right) = y - 1$

$\implies \frac{1}{2} x - \frac{1}{2} = y - 1$

$\implies \frac{1}{2} x + \frac{1}{2} = y$

That is the equation of the tangent line at the point $\left(1 , 1\right)$

A normal line is perpendicular to the tangent line.

We can find the slope of the normal line by finding the negative reciprocal of the slope of the tangent line.

$\frac{1}{2} \implies - 2$

We use the formula $m \left(x - {x}_{1}\right) = y - {y}_{1}$ once again.

$\implies - 2 \left(x - 1\right) = y - 1$

$\implies - 2 x + 2 = y - 1$

$\implies - 2 x + 3 = y$

That is the equation of the normal line at the point $\left(1 , 1\right)$