Question

How do you find the derivative of `y=xln^3x`?

Answer 1

`dy/dx=ln^2x(lnx+3)`

Explanation:

We need to start with the product rule. Where `u` and `v` are both functions, if `y=uv`, then:

`dy/dx=(du)/dxv+u(dv)/dx`

Thus, where `y=xln^3x`:

`dy/dx=(d/dxx)ln^3x+x(d/dxln^3x)`

Now we have two internal derivatives we need to figure out. The first is basic: `d/dxx=1`.

For the second derivative, we need the chain rule. First, note that we have a function cubed: `ln^3x=(lnx)^3`. Through the power rule, `d/dxx^3=3x^2`, but if there were a more complex function instead of `x` we see that `d/dxu^3=3u^2(du)/dx`--that is, we still do the power rule but then multiply by the derivative of the inner function.

Thus, `d/dxln^3x=3ln^2x(d/dxlnx)`.

`dy/dx=1*ln^3x+x(3ln^2x)(d/dxlnx)`

Recall that `d/dxlnx=1/x`:

`dy/dx=ln^3x+3xln^2x(1/x)`

`dy/dx=ln^3x+3ln^2x`

If you wish:

`dy/dx=ln^2x(lnx+3)`