How do you find the derivative of $f (x) = {log}_{x} 2$ $f(x) = log_x 2$ ?

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Answer 1 · 2016-04-24T16:21:01

Change the base: ${log}_{x} 2 = \frac{ln 2}{ln x}$ $log_x 2 = (ln2)/(lnx)$

$f (x) = {log}_{x} 2 = \frac{ln 2}{ln x}$ $f(x) = log_x 2 = (ln2)/(lnx)$

I think maybe the quotient rule is clearest rather than rewriting any more.

$f'(x) = ((0)(lnx)-(ln2)(1/x))/(lnx)^2$

$= -ln2/(x(lnx)^2)$

If wished, we can rewrite again:

$= -(ln2)/lnx 1/(xlnx)$

$= -log_x 2/(xlnx)$

How do you find the derivative of f(x) = log_x 2f(x)=logx2f(x) = log_x 2?