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

Jun 9, 2017

Equation of tangent is $9 y + \left({e}^{2} - 4\right) x = 5 {e}^{2} - 11$

#### Explanation:

The slope of the tangent is the same as the slope of curve at that point, which is given by the value of derivative of the function at that point.

Hence, slope of tangent to curve $f \left(x\right) = \frac{{e}^{x} - 1}{{x}^{2} - 1}$ at $x = 2$ is given by $f ' \left(2\right)$. And as the tangent passes through $\left(x , f \left(x\right)\right)$ i.e. $\left(2 , f \left(2\right)\right)$, the equation of tangent is

$y - f \left(2\right) = f ' \left(2\right) \left(x - 2\right)$

Now $f \left(2\right) = \frac{{e}^{2} - 1}{{2}^{2} - 1} = \frac{{e}^{2} - 1}{3}$

and using quotient rule $f ' \left(x\right) = \frac{\left({x}^{2} - 1\right) {e}^{x} - \left({e}^{x} - 1\right) 2 x}{{x}^{2} - 1} ^ 2$

= $\frac{{x}^{2} {e}^{x} - {e}^{x} - 2 x {e}^{x} + 2 x}{{x}^{2} - 1} ^ 2$

and slope at $f \left(2\right)$ is $f ' \left(2\right) = \frac{4 {e}^{2} - {e}^{2} - 4 {e}^{2} + 4}{9} = \frac{4 - {e}^{2}}{9}$

Hence, equation of tangent is

$y - \frac{{e}^{2} - 1}{3} = \frac{4 - {e}^{2}}{9} \left(x - 2\right)$

or $9 y - 3 {e}^{2} + 3 = 4 x - 8 - {e}^{2} x + 2 {e}^{2}$

or $9 y + \left({e}^{2} - 4\right) x = 5 {e}^{2} - 11$

graph{(9y+(e^2-4)x-5e^2+11)(x^2y-y-e^x+1)=0 [-4.785, 5.215, -1.18, 3.82]}