Differentiate the following function

y=x^3x
ln y = 3x ln x
(1/y)dy/dx =3 ln x +3 x(1/x)=3 (ln x + 1)
dy/dx=3x^3x (ln x +1) (ANS.)

y= x^3x
ln y=3x(lnx)
so (1/y)y'=3x/x+3lnx
so (1/y)y'=3+3lnx
y'=3(1+lnx)*y
=3(ln+1)(x^3x)

∵dlny/dx=dlny/dy*dy/dx (chain rule)

∴dy/dx = (dlny/dx) / (dlny/dy)

ln y = 3x ln x

∴dy/dx = (d3xlnx/dx) / (dlny/dy)

= (lnx d3x/dx + 3x dlnx/x) / (1/y)

= (3lnx + 3x*1/x) / (1/y)

= 3y(lnx + 1)

=3(x^3x)(lnx+1)