What is the equation of the line tangent to # f(x)=(x^2-x )e^(1-x^2)# at # x=1#?

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Answer 1 · 2016-04-28T15:35:58

#y=x-1#

Use Product Rule to find f'(x) then put in 1 for x to find the slope.
Let #f=x^2-x# and #g=e^(1-x^2)#
#f'=2x-1# and #g'=-2x e^(1-x^2)#

#f'(x)=fg'+gf'#

#f'(x)=(-2x^3+2x^2) e^(1-x^2) + (2x-1) e^(1-x^2)#

#f'(1)=(-2+2)e^(1-1)+(2-1)e^(1-1) = 0+1e^0=1#

Equation of tangent line
#y=f(a)+f'(a)(x-a)#

#y=f(1)+f'(1)(x-1)#

#f(1)=(1-1)e^(1-1)=0#

#y=0+1(x-1)#

#y=x-1#