[MOVED] Help with differentiation

[SamanthaMFR]

I have y=4x^3+(6/x)+3e^(2x)+ln2x

I think the answer is 12x^2-(6/x^2)+6e^(2x)+2/x

I think this is the right answer, but Im not sure with ln2x, is it 2/x?

 

[pka]

I have y=4x^3+(6/x)+3e^(2x)+ln2x
I think this is the right answer, but Im not sure with ln2x, is it 2/x?

No that answer is incorrect.
It should be \(\displaystyle \dfrac{1}{x}\) because \(\displaystyle \log(2x)=\log(2)+\log(x).\)

 

[HallsofIvy]

Equivalently, use the chain rule. Let u= 2x so ln(2x)= ln(u) and d(ln(u))/du= 1/u. But du/dx= d(2x)/dx= 2 so d(ln(x))/dx= [d(ln(u))/du][du/dx]= (1/u)(2)= (1/2x)(2)= 1/x.

Again, why is this posted in "Beginning Algebra"?