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!