# What is the equation of the tangent line of f(x) =e^x/lnx-x at x=4?

May 23, 2018

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

#### Explanation:

$f \left(x\right) = {e}^{x} / \ln x - x$, ${D}_{f} = \left(0 , 1\right) \cup \left(1 , + \infty\right)$

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

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

${e}^{x} / \ln x - {e}^{x} / \left(x {\ln}^{2} x\right) - 1$

The equation of the tangent line at $M \left(4 , f \left(4\right)\right)$ will be

$y - f \left(4\right) = f ' \left(4\right) \left(x - 4\right)$ $\iff$

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

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