# What is the equation of the line that is normal to f(x)= x/sqrt( 2x+2)  at  x=2 ?

Nov 15, 2017

$y = - \frac{3 \sqrt{6}}{2} x + \frac{10 \sqrt{6}}{3}$

#### Explanation:

First we need to differentiate the function in order to find the gradient at $x = 2$.

If we rewrite $\frac{x}{\sqrt{2 x + 2}}$ as $x \cdot {\left(2 x + 2\right)}^{- \frac{1}{2}}$ we can use the product rule and the chain rule.

The product rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(a \cdot b\right) = b \cdot \frac{\mathrm{da}}{\mathrm{dx}} + a \cdot \frac{\mathrm{db}}{\mathrm{dx}}$

$\frac{d}{\mathrm{dx}} \left(x\right) = 1$

$\frac{d}{\mathrm{dx}} {\left(2 x + 2\right)}^{- \frac{1}{2}} = - \frac{1}{2} {\left(2 x + 2\right)}^{- \frac{3}{2}} \cdot \left(2\right) = - {\left(2 x + 2\right)}^{- \frac{3}{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} x \cdot {\left(2 x + 2\right)}^{- \frac{1}{2}} = {\left(2 x + 2\right)}^{- \frac{1}{2}} + x \cdot \left(- {\left(2 x + 2\right)}^{- \frac{3}{2}}\right)$

->(1)/((2x+2)^(1/2))-(x)/(2x+2)^(3/2)=((2x+2)-x)/((2x+2)^(3/2))=(x+2)/((2x+2)^(3/2))=(x+2)/((2x+2)(2x+2)^(1/2))=1/2(x+2)/((x+1)sqrt(2x+2)

f^'(x)=1/2(x+2)/((x+1)sqrt(2x+2)

Gradient at $x = 2$

$\frac{1}{2} \frac{2 + 2}{\left(2 + 1\right) \sqrt{2 \left(2\right) + 2}} = \frac{4}{6 \sqrt{6}} = \frac{\sqrt{6}}{9}$

For the normal line:

$\frac{\sqrt{6}}{9} = - \frac{3 \sqrt{6}}{2}$

Point on the line:

Plugging 2 into original function:

$\frac{2}{\sqrt{2 \left(2\right) + 2}} = \frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3}$

$\left(2 \textcolor{w h i t e}{8} , \frac{\sqrt{6}}{3}\right)$

$y - \frac{\sqrt{6}}{3} = - \frac{3 \sqrt{6}}{2} \left(x - 2\right)$

$\to y = - \frac{3 \sqrt{6}}{2} x + \frac{6 \sqrt{6}}{2} + \frac{\sqrt{6}}{3} = - \frac{3 \sqrt{6}}{2} x + \frac{10 \sqrt{6}}{3}$

Normal line:

$y = = - \frac{3 \sqrt{6}}{2} x + \frac{10 \sqrt{6}}{3}$

Or:

$6 y = - 9 \sqrt{6} \cdot x + 20 \sqrt{6}$

Plot: