How do you differentiate #f(x)=ln((x^3-x ^2 -3x + 1) ^(2/5))# using the chain rule?

Question

Christopher Agresta · Accepted Answer

Working your way outwards you aim to end up with #f^'(x) = (dy)/dx# and you do it in stages. You then just multiply all the separate differentials together to obtain the final answer. ~~~~~~~~~~~~~The core of the function ~~~~~~~~~~~~~~~~
Let #u = x^3 -x^2-3x+1# Then #(du)/dx = 3x^2 -x -3# ~~~~~~~~~~~~~Out one stage ~~~~~~~~~~~~~~~~~~~~ Let #v = u^(2/5)# then #(dv)/(du) = 2/5 u^(-3/5)# ~~~~~~~~~~~~~~Out another stage ~~~~~~~~~~~~~~~~ Let #y = ln(v)# then #(dy)/(dv) = 1/v# ~~~~~~~~~~~~~~~Putting it all togther~~~~~~~~~~~~~~~~~~~~~ But #"  "(dy)/dx = (dy)/(dv) times (dv)/(du) times (du)/dx # When you cancel out the right hand side you do indeed end up with #(dy)/dx# . So the chain rule is just multiplying out the values obtained at each stage

Axel Alvarez · Answer

undefined To differentiate $ f(x) = \ln((x^3 - x^2 - 3x + 1)^{2/5}) $ using the chain rule, we first differentiate the outer function $\ln(u)$ with respect to $u$, then multiply by the derivative of the inner function $u$, where $u = (x^3 - x^2 - 3x + 1)^{2/5}$.

The derivative of $\ln(u)$ with respect to $u$ is $\frac{1}{u}$, and the derivative of the inner function $u$ with respect to $x$ is $\frac{2}{5}(x^3 - x^2 - 3x + 1)^{-3/5}(3x^2 - 2x - 3)$.

Thus, the derivative of $f(x)$ with respect to $x$ using the chain rule is:

$$f'(x) = \frac{1}{(x^3 - x^2 - 3x + 1)^{2/5}} \cdot \frac{2}{5}(x^3 - x^2 - 3x + 1)^{-3/5}(3x^2 - 2x - 3)$$