How do you differentiate #x^2sqrt(x^4+1)#?

Question

1169 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-05T09:25:36

#=(4x^5+2x)/(sqrt(x^4+1))#

Since

#f(x)=g(x)*h(x)->f'(x)=g'(x)*h(x)+g(x)*h'(x)#
and
#f(x)=sqrt(g(x))->f'(x)=1/(2sqrt(g(x)))*g'(x)#,

then

#y=x^2sqrt(x^4+1)->y'=2x*sqrt(x^4+1)+x^2*1/(cancel2sqrt(x^4+1))*cancel4^2x^3#

#=2x*sqrt(x^4+1)+(2x^5)/(sqrt(x^4+1))#

#=(2x*(x^4+1)+2x^5)/(sqrt(x^4+1))#

#=(4x^5+2x)/(sqrt(x^4+1))#