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

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Answer 1 · 2017-01-19T12:55:01

$(dy)/(dx)=1/(xlnb)$

Before differentiating we need to change the log to base e

$y=log_bx$

$=>b^y=x$

take logs to base e

$lnb^y=lnx$

$=>ylnb=lnx$

now differentiate wrt $x$

$d/(dx)(ylnb)=d/(dx)(lnx)$

$lnb(dy)/(dx)=1/x$

$(dy)/(dx)=1/(xlnb)$

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