Derivative of 2cos(2x)/sin(2x)

Derivative of 2cos(2x)/sin(2x). Simple step by step solution, to learn. Simple, and easy to understand, so don`t hesitate to use it as a solution of your homework.

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Derivative of 2cos(2x)/sin(2x): 


((2*cos(2*x))/sin(2*x))'

((2*cos(2*x))'*sin(2*x)-(2*cos(2*x)*(sin(2*x))'))/((sin(2*x))^2)

(((2)'*cos(2*x)+2*(cos(2*x))')*sin(2*x)-(2*cos(2*x)*(sin(2*x))'))/((sin(2*x))^2)

((0*cos(2*x)+2*(cos(2*x))')*sin(2*x)-(2*cos(2*x)*(sin(2*x))'))/((sin(2*x))^2)

((0*cos(2*x)+2*-sin(2*x)*(2*x)')*sin(2*x)-(2*cos(2*x)*(sin(2*x))'))/((sin(2*x))^2)

((0*cos(2*x)+2*-sin(2*x)*((2)'*x+2*(x)'))*sin(2*x)-(2*cos(2*x)*(sin(2*x))'))/((sin(2*x))^2)

((0*cos(2*x)+2*-sin(2*x)*(0*x+2*(x)'))*sin(2*x)-(2*cos(2*x)*(sin(2*x))'))/((sin(2*x))^2)

((0*cos(2*x)+2*-sin(2*x)*(0*x+2*1))*sin(2*x)-(2*cos(2*x)*(sin(2*x))'))/((sin(2*x))^2)

((0*cos(2*x)+2*2*(-sin(2*x)))*sin(2*x)-(2*cos(2*x)*(sin(2*x))'))/((sin(2*x))^2)

((0*cos(2*x)+2*-2*sin(2*x))*sin(2*x)-(2*cos(2*x)*(sin(2*x))'))/((sin(2*x))^2)

(-4*sin(2*x)*sin(2*x)-(2*cos(2*x)*(sin(2*x))'))/((sin(2*x))^2)

(-4*sin(2*x)*sin(2*x)-(2*cos(2*x)*cos(2*x)*(2*x)'))/((sin(2*x))^2)

(-4*sin(2*x)*sin(2*x)-(2*cos(2*x)*cos(2*x)*((2)'*x+2*(x)')))/((sin(2*x))^2)

(-4*sin(2*x)*sin(2*x)-(2*cos(2*x)*cos(2*x)*(0*x+2*(x)')))/((sin(2*x))^2)

(-4*sin(2*x)*sin(2*x)-(2*cos(2*x)*cos(2*x)*(0*x+2*1)))/((sin(2*x))^2)

(-4*sin(2*x)*sin(2*x)-(2*cos(2*x)*2*cos(2*x)))/((sin(2*x))^2)

(-4*(sin(2*x))^2-(4*(cos(2*x))^2))/((sin(2*x))^2)

The calculation above is a derivative of the function f (x)

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