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

Question

1559 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-07T19:39:50

#y=-13.24015x+44.6855#

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

#y=f(a)-1/(f'(a))(x-a)#

#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=-13.24015x+44.6855#