# What is the equation of the normal line of f(x)=x^2/(4x-1) at x=4?

Feb 14, 2016

$y = \frac{225}{8} x - \frac{2233}{20}$

#### Explanation:

To find the equation of the normal , require to find the gradient of the tangent (m) and (a , b ) a point on the line. Evaluating f'(4) will provide m , and evaluating (f(4) will give (a , b ).

differentiating f(x) using the $\textcolor{b l u e}{\text{ quotient rule }}$

If fx) $= g \frac{x}{h \left(x\right)}$
then f'(x) =( h(x).g'(x) - g(x).h'(x))/( (h(x))^2

hence f'(x) =$\frac{\left(4 x - 1\right) \frac{d}{\mathrm{dx}} \left({x}^{2}\right) - {x}^{2} \frac{d}{\mathrm{dx}} \left(4 x - 1\right)}{4 x - 1} ^ 2$

$= \frac{\left(4 x - 1\right) .2 x - {x}^{2} .4}{4 x - 1} ^ 2 = \frac{8 {x}^{2} - 2 x - 8 {x}^{2}}{4 x - 1} ^ 2$

hence f'(x)$= - \frac{2 x}{4 x - 1} ^ 2$
and f'(4) = $- \frac{8}{225}$

now f(4) = $\frac{16}{15} \Rightarrow \left(a , b\right) = \left(4 , \frac{16}{15}\right)$
The product of the gradients of the tangent and the normal equal -1.

so m of normal $\times - \frac{8}{225} = - 1 \textcolor{b l a c k}{\text{ m normal }} = \frac{225}{8}$

equation of normal : y - b = m(x - a)

y - $\frac{16}{15} = \frac{225}{8} \left(x - 4\right)$

$\Rightarrow y = \frac{225}{8} x - \frac{2233}{20}$