Changing convention from #f(x)# to #y# to calculate equation of tangent: #y=(x^2-x^3)e^(2x^2-x)#
Let #f=x^2-x^3# and #u=2x^2-x# and #g=e^(2x^2-x)=e^u#
#f'=2x-3x^2#
#u'=4x-1#
#g'=(dg)/(du)(du)/dx=e^u(4x-1)=(4x-1)e^(2x^2-x)#
#y'=f'g+g'f=(2x-3x^2)e^(2x^2-x)+(x^2-x^3)(4x-1)e^(2x^2-x)#
#y'(ln5)=(2(ln5)-3(ln5)^2)e^(2(ln5)^2-ln5)+((ln5)^2-(ln5)^3)(4ln5-1)e^(2(ln5)^2-ln5)#
#=-467.1# (4s.f.)
#y(ln5)=((ln5)^2-(ln5)^3)e^(2(ln5)^2-ln5)=-56.13# (4s.f.)
#y-y_1=y'(x-x_1)#
#y+56.13=-467.1(x-ln5)#
#y=751.8-467.1x# (4s.f.)
Checking on graph:
graph{y=(x^2-x^3)e^(2x^2-x) [1.5, 1.7, -57.852, -54.594]}
At #x=ln5~~1.61#, #y~~56.1# and the gradient is very steep (step of .01 right #~~#4.5 down, so the above solution is correct.
