What is the equation of the normal line of f(x)=x/lnx at x=4 ?

Feb 17, 2016

$y = \frac{\ln \left(4\right) - 1}{\ln {\left(4\right)}^{2}} \left(x - 4\right) + \frac{4}{\ln 4}$

Explanation:

First of all, we need a point of intersection. We know the line will intersect $f \left(x\right)$ at $\left(4 , f \left(4\right)\right)$ so we will use to obtain the point.

$f \left(4\right) = \frac{4}{\ln} \left(4\right)$

So we know at least $\left(4 , \frac{4}{\ln} 4\right)$ is a point on the line.

Next we need the gradient of the line and this is obtained by taking the derivative of $f \left(x\right)$. Using the quotient rule to differentiate we get:

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

Now find $f ' \left(4\right) = \frac{\ln \left(4\right) - 1}{\ln} {\left(4\right)}^{2}$

Thus we have the gradient of the line.

Now substitute into the formula:

$y - b = m \left(x - a\right)$ where $m$ is the gradient and $\left(a , b\right)$ is the known point of intersection:

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

$y = \frac{\ln \left(4\right) - 1}{\ln {\left(4\right)}^{2}} \left(x - 4\right) + \frac{4}{\ln 4}$

We can better visualise this from the graphs of the function: