# What is the equation of the tangent line of f(x) =sqrt(x+15)-x/(x-15) at x=5?

Apr 15, 2018

$20 y - \left(3 + \sqrt{5}\right) x + 5 - 35 \sqrt{5} = 0$

#### Explanation:

$y = \sqrt{x + 15} - \frac{x}{x - 15}$

Substitute for $x$ value to get the coordinates of the point which lies on the tangent

$y = \frac{1 + 4 \sqrt{5}}{2}$

Differentiate the function Using

$y ' = \frac{1}{2 \sqrt{x + 15}} - \frac{\left(x - 15\right) \times 1 - x \times 1}{x - 15} ^ 2$

Simplify

$y ' = \frac{1}{2 \sqrt{x + 15}} + \frac{15}{x - 15} ^ 2$

Now We substitute for x=5 to get the Slope of the Tangent $m$

$y {'}_{x = 5} = \frac{3 + \sqrt{5}}{20}$=$m$

now Substitute in the straight line equation

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - \frac{1 + 4 \sqrt{5}}{2} = \frac{3 + \sqrt{5}}{20} \left(x - 5\right)$

$20 y - \left(3 + \sqrt{5}\right) x + 5 - 35 \sqrt{5} = 0$