Question #e21ad

1 Answer
Jun 18, 2017

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

Explanation:

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

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

Then take the derivative

d/dx[log_2(3x-cos(2x))]=d/dx[(log(3x-cos(2x)))/(log(2))]ddx[log2(3xcos(2x))]=ddx[log(3xcos(2x))log(2)]

Factor out the constant

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

Use the Chain Rule on the derivative of the log function

=1/log(2)xx1/(3x-cos(2x))(d/dx[3x-cos(2x)])=1log(2)×13xcos(2x)(ddx[3xcos(2x)])

=1/log(2)xx1/(3x-cos(2x))xx(d/dx[3x]-d/dx[cos(2x)])=1log(2)×13xcos(2x)×(ddx[3x]ddx[cos(2x)])

=1/log(2)xx1/(3x-cos(2x))xx(3-(-sin(2x)))=1log(2)×13xcos(2x)×(3(sin(2x)))

=1/log(2)xx1/(3x-cos(2x))xx(3+sin(2x))=1log(2)×13xcos(2x)×(3+sin(2x))

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