# What is the equation of the tangent line of f(x) =(4x) / (3-4x^2-x)  at  x = 1?

Jan 20, 2017

$y = 7 x - 9$

#### Explanation:

$y = k x + n$, where $k = f ' \left({x}_{0}\right) , {x}_{0} = 1$

Lets find the first derivative (using quotient rule):

$f ' \left(x\right) = \frac{4 \left(3 - 4 {x}^{2} - x\right) - 4 x \left(- 8 x - 1\right)}{3 - 4 {x}^{2} - x} ^ 2$

$f ' \left(x\right) = \frac{12 - 16 {x}^{2} - 4 x + 32 {x}^{2} + 4 x}{3 - 4 {x}^{2} - x} ^ 2$

$f ' \left(x\right) = \frac{16 {x}^{2} + 12}{3 - 4 {x}^{2} - x} ^ 2$

$k = f ' \left({x}_{0}\right) = f ' \left(1\right) = \frac{16 + 12}{3 - 4 - 1} ^ 2 = \frac{28}{- 2} ^ 2 = 7$

For ${x}_{0} = 1 \implies {y}_{0} = f \left(1\right) = \frac{4}{3 - 4 - 1} = \frac{4}{- 2} = - 2$

${y}_{0} = k {x}_{0} + n \implies - 2 = 7 \cdot 1 + n \implies n = - 9$

Finally,

$y = 7 x - 9$