# What is the equation of the normal line of f(x)=sqrt(x/(x+1)  at x=4 ?

Jun 28, 2016

$100 x + \sqrt{5} y - 402 = 0.$

#### Explanation:

Let $y = f \left(x\right) = \sqrt{\frac{x}{x + 1}} ,$ so, $x = 4 \Rightarrow y = f \left(4\right) = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5.}}$

So, we reqiure the eqn. of normal to the curve : $y = f \left(x\right) = \sqrt{\frac{x}{x + 1}} ,$ at the pt. $\left(4 , \frac{2}{\sqrt{5}}\right)$

We recall that ${\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{x = 4 , y = \frac{2}{\sqrt{5}}} = f ' \left(4\right)$ gives us the slope tgt. line to the given curve.

Now, to diff. y, we can use the Quotient Rule. Instead, have a look at these two methods :-

METHOD I :-

Taking log. of both sides of the given eqn, we get, $\ln y = \frac{1}{2} \left\{\ln x - \ln \left(x + 1\right)\right\} \Rightarrow \frac{d}{\mathrm{dx}} \left(\ln y\right) = \frac{1}{2} \frac{d}{\mathrm{dx}} \left\{\ln x - \ln \left(x + 1\right)\right\} \Rightarrow \frac{d}{\mathrm{dy}} \left(\ln y\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} \left\{\frac{1}{x} - \frac{1}{x + 1}\right\} = \frac{1}{2 x \left(x + 1\right)} \Rightarrow \frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 x \left(x + 1\right)} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{2 x \left(x + 1\right)}$

$\therefore {\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{x = 4 , y = \frac{2}{\sqrt{5}}} = f ' \left(4\right) = \frac{\frac{2}{\sqrt{5}}}{2 \cdot 4 \cdot 5} = \frac{1}{20 \sqrt{5}} .$

METHOD II :-

We write the eqn. of the given curve : ${y}^{2} \left(x + 1\right) = x \Rightarrow \frac{d}{\mathrm{dx}} {y}^{2} \left(x + 1\right) = \frac{d}{\mathrm{dx}} x \Rightarrow {y}^{2} \cdot \frac{d}{\mathrm{dx}} \left(x + 1\right) + \left(x + 1\right) \cdot \frac{d}{\mathrm{dx}} {y}^{2} = 1 \Rightarrow {y}^{2} + \left(x + 1\right) \left\{\frac{d}{\mathrm{dy}} {y}^{2}\right\} \frac{\mathrm{dy}}{\mathrm{dx}} = 1 \Rightarrow {y}^{2} + 2 y \left(x + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1 \Rightarrow \frac{x}{x + 1} + 2 y \left(x + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1. \ldots . . , \left[a s , {y}^{2} = \frac{x}{x + 1}\right] \Rightarrow 2 y \left(x + 1\right) \frac{\mathrm{dy}}{\mathrm{dx}} = 1 - \frac{x}{x + 1} = \frac{1}{x + 1} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 y {\left(x + 1\right)}^{2}}$

${\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{x = 4 , y = \frac{2}{\sqrt{5}}} = f ' \left(4\right) = \frac{1}{2 \cdot \left(\frac{2}{\sqrt{5}}\right) \cdot 25} = \frac{1}{20 \sqrt{5}} ,$ as before!

Since normal line is perp. to tgt., its slope is $- \frac{1}{f ' \left(4\right)} = - 20 \sqrt{5} , .$ and it passes thro. pt. $\left(4 , \frac{2}{\sqrt{5}}\right) ,$ its eqn. is $: y - \frac{2}{\sqrt{5}} = - 20 \sqrt{5} \left(x - 4\right)$ or, $\sqrt{5} y - 2 + 100 x - 400 = 0 ,$ i.e., $100 x + \sqrt{5} y - 402 = 0.$