#f(x)=((5−x)(4−x^2))/(x^3−1);quadquadquada=3#
normal line: #quady=f(a)-1/(f'(a))(x-a)#
#color(blue)(f(3)=)((5−3)(4−(3)^2))/(3^3−1)=(2*(-5))/(26)=color(blue)(-5/13)~~-0.384#
Let: #g(x)=(5−x)(4−x^2)#
Then: #color(red)(g'(x)=)-1(4-x^2)+(5-x)*(-2x)=color(red)(-(4-x^2)-2x(5-x))#
#f'(x)=([-(4-x^2)-2x(5-x)]xx(x^3−1)-(5−x)(4−x^2)xx3x^2)/(x^3−1)^2#
Don't bother to simplify. Just substitute x for 3
#f'(3)=([-(4-9)-6(2)]xx(26)-(2)(-5)xx27)/(26)^2#
#f'(3)=([+5-12]xx(26)+10xx27)/(26)^2=(-7xx26+270)/26^2#
#color(orange)(f'(3)=)(-182+270)/676=88/676=44/338=color(orange)(22/169)~~0.13#
#y=-5/13-1/(22/169)(x-3)#
#y=-169/22x+6481/286#
#y=-7.6bar(81)x+22.66#
