How do you differentiate $log_3 (2x) / x^2$ ?

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Answer 1 · 2016-11-07T12:10:06

$(1-2ln(2x))/(x^3ln(3))$

The first step is to write $log_3(2x)$ as $ln(2x)/ln(3)$ through the change of base formula.

$y=log_3(2x)/x^2=ln(2x)/(ln(3)x^2)$

$ln(3)$ is just a constant, don't forget this:

$y=1/ln(3) ln(2x)/x^2$

When differentiating this, use the quotient rule:

$dy/dx=1/ln(3)((d/dxln(2x))x^2-ln(2x)(d/dxx^2))/(x^2)^2$

These derivatives are:

This required the chain rule. The following is just the power rule:

So:

$dy/dx=1/ln(3)(1/x(x^2)-ln(2x)*2x)/x^4$

$dy/dx=1/ln(3)(x-2xln(2x))/x^4$

$dy/dx=1/ln(3)(1-2ln(2x))/x^3$

$dy/dx=(1-2ln(2x))/(x^3ln(3))$

How do you differentiate log_3 (2x) / x^2log_3 (2x) / x^2?