What is the equation of the tangent line of #f(x)=xlnx^2-x^2lnx# at #x=3#?

Question

1314 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-06T08:25:10

#y=-[1+4ln(3)]x+3[1+3ln(3)]#

#f(x)=xln(x^2)-x^2ln(x)#

#=(2-x)xln(x)#

The equation of the tangent line #x=3# is the same as the first order taylor expansion of #f# about #x=3#.

#f(x)~~f(3)+f'(3)(x-3)=y#

#f(3)=(2-3)(3)ln(3)#

#=-3ln(3)#

#f'(x)=frac{d}{dx}((2-x)xln(x))#

#=(2-x)xfrac{d}{dx}(ln(x))+(2-x)frac{d}{dx}(x)ln(x)+frac{d}{dx}(2-x)xln(x)#

#=(2-x)xfrac{1}{x}+(2-x)(1)ln(x)+(-1)xln(x)#

#=2-x+2(1-x)ln(x)#

#f'(3)=2-3+2(1-3)ln(3)#

#=-1-4ln(3)#

#y=-(1+4ln(3))(x-3)-3ln(3)#

#=-(1+4ln(3))x+9ln(3)+3#