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

Question

Aurora Alvarado · Accepted Answer

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#

Scarlett Allison · Answer

undefined To find the derivative of $ s = \sqrt{t} $ using logarithmic differentiation, follow these steps:

1. Take the natural logarithm of both sides of the equation: $ \ln(s) = \ln(\sqrt{t}) $.
2. Use the properties of logarithms to simplify the right side of the equation: $ \ln(s) = \frac{1}{2}\ln(t) $.
3. Differentiate both sides of the equation with respect to $ t $.
4. Apply the chain rule and the derivative of the natural logarithm.
5. Solve for $ \frac{ds}{dt} $.

The result will be $ \frac{ds}{dt} = \frac{1}{2s} $.