How do you differentiate $y=log_x(2)$ ?

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Answer 1 · 2016-10-25T18:11:49

$x^y = 2$

$ln(x^y) = ln2$

$yln(x) = ln2$

By the product rule and the differential identity $(ln(x))' = 1/x$ :

$1(dy/dx)lnx + y(1/x) = 0$

$lnx(dy/dx) = -y/x$

$dy/dx = (-y/x)/lnx$

$dy/dx = -y/(xlnx)$

Since $y = ln_x(2)$ :

$dy/dx =- (ln_x(2))/(xlnx)$

We can simplify this further using the rule $log_a(n) = logn/loga$ .

$dy/dx = -(ln2/lnx)/(xlnx)$

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

Hopefully this helps!

How do you differentiate y=log_x(2)y=log_x(2)?