Question

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

Answer 1

`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.