Question

How do you differentiate `y=x^2ln(2x)+xln(3x)+4lnx`?

Answer 1

`y' = (2x^2ln(2x) + x^2 + xln(3x) + x + 4)/x`

Explanation:

Use the product and chain rules.

Developing a formula for the derivative of `ln(ax)`, where `a` is a constant

By the chain rule, letting `y = ln u` and `u =ax`. Then `du = a dx` and `dy = 1/u du`. We have:

`dy/dx= 1/u * a`

`dy/dx = a/(ax)`

`dy/dx = 1/x`

Apply the formula to the problem

`y' = 2xln(2x) + x^2(1/x) + ln(3x) + x(1/x) + 4/x`

`y' = 2xln(2x) + x + ln(3x) + 1 + 4/x`

`y' = (2x^2ln(2x) + x^2 + xln(3x) + x + 4)/x`

Hopefully this helps!