How do you differentiate $y= log _a x$ ?

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Answer 1 · 2016-07-22T18:14:33

$y' = 1/(xln a)$

the easiest way is to shift the base to e

so $y = log_a x = (log_e x)/(log_e a)$ {small demo of what that is so is set out below}

thusly

$y' = 1/x *1/(log_e a) = 1/(xln a)$

the demo

$y = log_a x implies a^y = x$ by definition

so we choose to use natural logs because they work so well with calculus

$ln a^y = ln x$

$y ln a = ln x$

$y = ( ln x)/(ln a)$

How do you differentiate y= log _a xy= log _a x?