Question

How do you find the derivative of `log _ 3 (2x) / x^2`?

Answer 1

`f'(x)=(1-2ln(2x))/(x^3ln(3))`

Explanation:

First, rewrite the logarithm in terms of the natural logarithm, since we know how to differentiate the natural logarithm.

It can be rewritten using the change of base formula: `log_3(2x)=ln(2x)/ln(3)`

Thus, the function can be written as

`f(x)=log_3(2x)/x^2=1/x^2ln(2x)/ln(3)=ln(2x)/(x^2ln(3))`

To differentiate this, use the quotient rule.

`f'(x)=(x^2ln(3)d/dx[ln(2x)]-ln(2x)d/dx[x^2ln(3)])/(x^2ln(3))^2`

We can find each of these derivatives individually:

The first requires the chain rule:

`d/dx[ln(2x)]=1/(2x)*d/dx[2x]=2/(2x)=1/x`

This can also be found through first splitting up `ln(2x)=ln(2)+ln(x)`...

`d/dx[ln(2)+ln(x)]=d/dx[ln(x)]=1/x`

To find the second, just use the power rule and remember that `ln(3)` is just a constant.

`d/dx[x^2ln(3)]=2xln(3)`

Plugging both these back in yields

`f'(x)=(x^2ln(3)(1/x)-ln(2x)(2xln(3)))/(x^4ln^2(3))`

Simplify.

`f'(x)=(xln(3)-2xln(2x)ln(3))/(x^4ln^2(3))`

`f'(x)=(xln(3)(1-2ln(2x)))/(x^4ln^2(3))`

`f'(x)=(1-2ln(2x))/(x^3ln(3))`