Question #7951d

1 Answer
Oct 11, 2016

#(tanx-xln(2x)sec^2x)/(xtan^2x)#

Explanation:

You just need to make use of the quotient rule and the chain rule.

Quotient Rule:

#d/dx[f(x)/g(x)]=(g(x)f'(x)-f(x)g'(x))/(g(x))^2#

Chain Rule:

#d/dx[f(g(x))]=f'(g(x))*g'(x)#
(basically, you get the derivative of #f'(x)# then multiply by the derivative of #g(x)#)

Solution:

#d/dx[ln(2x)/tanx]#

By quotient rule:

#[1]" "=(tanx*D_x[ln(2x)]-ln(2x)*D_x[tanx])/tan^2x#

The derivative of #tanx# is #sec^2x:#

#[2]" "=(tanx*D_x[ln(2x)]-ln(2x)*sec^2x)/tan^2x#

To get the derivative of #ln(2x)#, you use chain rule. The derivative of #ln(x)# is #1/x#

#[3]" "=(tanx*1/(2x)*D_x[2x]-ln(2x)*sec^2x)/tan^2x#

#[4]" "=(tanx*1/(2x)*2-ln(2x)*sec^2x)/tan^2x#

#[5]" "=(tanx*1/x-ln(2x)*sec^2x)/tan^2x*x/x#

#[6]" "=color(red)((tanx-xln(2x)sec^2x)/(xtan^2x))#