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

Jan 7, 2018

$y = - 13.24015 x + 44.6855$

#### Explanation:

f(x)=sqrt(e^(2x)−x^2)/(x−1)^2;quadquadquadquada=3

$y = f \left(a\right) - \frac{1}{f ' \left(a\right)} \left(x - a\right)$

f(3)=sqrt(e^(6)−9)/4

f'(x)=(1/2(e^(2x)−x^2)^(-1/2)*(2e^(2x)-2x)(x-1)^2-sqrt(e^(2x)−x^2)(2(x-1)))/(x−1)^4

f'(x)=(1/2(e^(2x)−x^2)^(-1/2)*(2e^(2x)-2x)(x-1)-2sqrt(e^(2x)−x^2))/(x−1)^3

f'(3)=((e^6−9)^(-1/2)*(2e^6-6)-2sqrt(e^6−9))/8~~0.0755

y=sqrt(e^(6)−9)/4-(8(x-3))/((e^6−9)^(-1/2)*(2e^6-6)-2sqrt(e^6−9))

$y = - 13.24015 x + 44.6855$