Question #7951d

1 Answer
Oct 11, 2016

(tanx-xln(2x)sec^2x)/(xtan^2x)tanxxln(2x)sec2xxtan2x

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))