How do you find the derivative of $y=log_4 2x$ ?

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Answer 1 · 2015-12-25T19:02:51

$y'=1/(xln4)$

Rewrite $y$ using the change of base formula.

$y=ln(2x)/ln4$

$y'=1/ln4*d/dx(ln(2x))$

Recall that $1/ln4$ is just a constant.

To find $d/dx=(ln(2x))$ , use the chain rule.

$ln(u)=1/u*(du)/dx$

Thus,

$d/dx(ln(2x))=1/(2x)d/dx(2x)=2/(2x)=1/x$

Plug this back into the $y'$ expression.

$y'=1/(xln4)$

How do you find the derivative of y=log_4 2xy=log_4 2x?