Question

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

Answer 1

`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 )`