What is the equation of the tangent line of #f(x) =sqrt(x^2e^x)# at #x=3#?

Question

1212 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-24T13:32:20

#y=11.2x-20.2#

Or

#y=(5e^(3/2))/2x-2e^(3/2)#
#y=e^(3/2)((5x)/2-2)#

We have:
#f(x)=(x^2e^x)^(1/2)#

#f'(x)=(x^2e^x)^(-1/2)/2*d/dx[x^2e^x]#

#f'(x)=(x^2e^x)^(-1/2)/2*(2xe^x+x^2e^x)#

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

#f'(x)=(2xe^x+x^2e^x)/(2(x^2e^x)^(1/2))=(2xe^x+x^2e^x)/(2sqrt(x^2e^x))#

#f'(3)=(2(3)e^3+3^2e^3)/(2sqrt(3^2e^3))=(5e^(3/2))/2~~11.2#

#y=mx+c#

#f(3)=sqrt(9e^3)=3e^(3/2)~~13.4#

#13.4=11.2(3)+c#

#c=13.4-11.2(3)=-20.2#

#y=11.2x-20.2#

Or

#y=(5e^(3/2))/2x-2e^(3/2)#
#y=e^(3/2)((5x)/2-2)#