Question

How do you differentiate `f(x)=lnx+ln2x+3x` using the sum rule?

Answer 1

`f'(x)=2/x+3`

Explanation:

The derivative of `lnx` is `1/x`.

To find the derivative of `ln2x` you'll need either the chain rule or the property of logarithms: `ln2x = ln2+lnx` and `ln2` is a constant, so its derivative is `0`.
Therefore, `d/dx(ln2x) = d/dx(ln2+lnx) = d/dx(ln2)+d/dx(lnx)=0+1/x=1/x`.
(Note that this is a use of the sum rule.)

The derivative of `3x` is `3`.

`f'(x) = d/dx(lnx)+d/dx(ln2x)+d/dx(3x)`
(This is the sum rule.)

`= 1/x+1/x+3 = 2/x+3`

If you prefer to write the derivative as a single ratio, get a common denominator and write:

`f'(x)=2/x+(3x)/x=(2+3x)/x " or " (3x+2)/x`