Question #e21ad

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Answer 1 · 2017-06-18T15:15:53

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

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

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

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

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

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

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