How do you differentiate #y=(x^2lnx)^4#?

1 Answer
Oct 30, 2016

# dy/dx = 4x(x^2lnx)^3 ( 1 + 2lnx ) #

Explanation:

# y = (x^2lnx)^4 #

By the chain rule,
# d/dxf(g(x)) =f'(g(x))g'(x)# or,# dy/dx=dy/(du)(du)/dx #,
we have:

# dy/dx = 4(x^2lnx)^3 d/dx(x^2lnx)#

Next, we need to use the product rule;
# d/dx(uv)=u(dv)/dx+v(du)/dx #
to get:

# dy/dx = 4(x^2lnx)^3 { (x^2)(d/dxlnx) + (d/dxx^2)(lnx) } #
# :. dy/dx = 4(x^2lnx)^3 { (x^2)(1/x) + (2x)(lnx) } #
# :. dy/dx = 4(x^2lnx)^3 ( x + 2xlnx ) #
# :. dy/dx = 4x(x^2lnx)^3 ( 1 + 2lnx ) #