How do you differentiate #f(x)=1/(x^sqrt(x-3))# using the chain rule?

1 Answer
Nov 19, 2015

Answer:

#dy/dx = -x^(-sqrt(x-3))(sqrt(x-3)/x +ln(x)/(2sqrt(x-3)))#

Explanation:

So we have

#y = x^(-sqrt(x-3))#

Taking the log of both sides we have

#ln(y) = -sqrt(x-3)*ln(x)#

Derivating we have

#d/dxln(y) = -sqrt(x-3)d/dxln(x) - ln(x)d/dxsqrt(x-3)#

Using the chain rule on the left hand side we have

#1/y*dy/dx = -sqrt(x-3)d/dxln(x) - ln(x)d/dxsqrt(x-3)#

Isolating #dy/dx# and solving the other, simple derivatives, we have

#dy/dx = -sqrt(x-3)*1/x*x^(-sqrt(x-3)) - ln(x)*(1/(2sqrt(x-3)))*x^(-sqrt(x-3))#

#dy/dx = -x^(-sqrt(x-3))(sqrt(x-3)/x +ln(x)/(2sqrt(x-3)))#

It's important to notice we can't simply say #-sqrt(x-3) = u# and do

#dy/dx = d/(du)x^u(du)/dx = ux^(u-1)(du)/dx#

Because that only works for constant values on the exponent! Nor can we say

#dy/dx = d/(du)x^u(du)/dx = ln(x)x^u(du)/dx#

Because that only works when the base is a constant! If both base and exponent change, you need to use implicit differentiation with logs.