How do you find the first and second derivative of #x^2 lnx#?

Question

5740 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-31T20:29:10

#(d y)/(d x)=2x*l n(x)+x" 'The first derivative '"#
#(d^2y)/(d x^2)=2*l n(x)+3" 'The second derivative '"#

#y=x^2* l n (x)#

#u=x^2" ; " u'=2x#

#v=l n (x)" ; "v^'=1/x#

#y=u*v#

#(d y)/(d x)=u^'*v+v^'*u#

#(d y)/(d x)=2x* l n(x)+1/cancel(x)*cancel(x)^2#

#(d y)/(d x)=2x*l n(x)+x" 'The first derivative '"#

#a=2x" ; "a^'=2#
#b=l n(x)" ; "b^'=1/x#

#(d^2y)/(d x^2)=a^'*b+b^'*a+1#

#(d^2y)/(d x^2)=2*l n(x)+1/cancel(x)*2cancel(x)+1#

#(d^2y)/(d x^2)=2*l n(x)+2+1#

#(d^2y)/(d x^2)=2*l n(x)+3" 'The second derivative '"#