Question #e21ad

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Question #e21ad

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Answer 1 · 2017-06-18T15:15:53

$\frac{d y}{d x} = \frac{3 + sin (2 x)}{log (2) (3 x - cos (2 x))}$ $dy/dx=(3+sin(2x))/(log(2)(3x-cos(2x)))$

Explanation:

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

${log}_{a} (b) = \frac{log (b)}{log (a)}$ $log_a(b)=log(b)/log(a)$

Then take the derivative

$\frac{d}{d x} [{log}_{2} (3 x - cos (2 x))] = \frac{d}{d x} [\frac{log (3 x - cos (2 x))}{log (2)}]$ $d/dx[log_2(3x-cos(2x))]=d/dx[(log(3x-cos(2x)))/(log(2))]$

Factor out the constant

$= \frac{1}{log (2)} \frac{d}{d x} [log (3 x - cos (2 x))]$ $=1/log(2) d/dx[log(3x-cos(2x))]$

Use the Chain Rule on the derivative of the log function

$= \frac{1}{log (2)} \times \frac{1}{3 x - cos (2 x)} (\frac{d}{d x} [3 x - cos (2 x)])$ $=1/log(2)xx1/(3x-cos(2x))(d/dx[3x-cos(2x)])$

$= \frac{1}{log (2)} \times \frac{1}{3 x - cos (2 x)} \times (\frac{d}{d x} [3 x] - \frac{d}{d x} [cos (2 x)])$ $=1/log(2)xx1/(3x-cos(2x))xx(d/dx[3x]-d/dx[cos(2x)])$

$= \frac{1}{log (2)} \times \frac{1}{3 x - cos (2 x)} \times (3 - (- sin (2 x)))$ $=1/log(2)xx1/(3x-cos(2x))xx(3-(-sin(2x)))$

$= \frac{1}{log (2)} \times \frac{1}{3 x - cos (2 x)} \times (3 + sin (2 x))$ $=1/log(2)xx1/(3x-cos(2x))xx(3+sin(2x))$

$= \frac{3 + sin (2 x)}{log (2) (3 x - cos (2 x))}$ $=(3+sin(2x))/(log(2)(3x-cos(2x)))$

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Question #e21ad

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