Question

Question #e21ad

Answer 1

`dy/dx=(3+sin(2x))/(log(2)(3x-cos(2x)))`

Explanation:

First, use the "change of base" for logarithms formula

`log_a(b)=log(b)/log(a)`

Then take the derivative

`d/dx[log_2(3x-cos(2x))]=d/dx[(log(3x-cos(2x)))/(log(2))]`

Factor out the constant

`=1/log(2) d/dx[log(3x-cos(2x))]`

Use the Chain Rule on the derivative of the log function

`=1/log(2)xx1/(3x-cos(2x))(d/dx[3x-cos(2x)])`

`=1/log(2)xx1/(3x-cos(2x))xx(d/dx[3x]-d/dx[cos(2x)])`

`=1/log(2)xx1/(3x-cos(2x))xx(3-(-sin(2x)))`

`=1/log(2)xx1/(3x-cos(2x))xx(3+sin(2x))`

`=(3+sin(2x))/(log(2)(3x-cos(2x)))`