How do you find the derivatives of #s=roott(t)# by logarithmic differentiation?

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Answer 1 · 2017-03-15T19:06:16

Please see the explanation.

Given: #s = root(t)(t)#

Rewrite as:

#s=t^(1/t)#

Use the natural logarithm on both sides:

#ln(s)=ln(t^(1/t))#

Use the property #ln(a^c) = (c)ln(a)#

#ln(s)=(1/t)ln(t)#

Differentiate both sides:

The left side is trivial but we need the quotient rule, #(g/h)'=(g'h-gh')/h^2# for the right side:

#g=ln(t)#
#h= t#
#g'=1/t#
#h'=1#

#(1/t)ln(t) = ((1/t)(t)-ln(t)(1))/t^2=(1-ln(t))/t^2#

#1/s(ds)/(dt) = (1-ln(t))/t^2#

Multiply both sides by s:

#(ds)/(dt) = (s(1-ln(t)))/t^2#

Substitute for s:

#(ds)/(dt) = (root(t)(t)(1-ln(t)))/t^2#