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

Question

1368 views around the world

You can reuse this answer
Creative Commons License

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))#