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

Oct 30, 2015

Working your way outwards you aim to end up with ${f}^{'} \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}}$ 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 $u = {x}^{3} - {x}^{2} - 3 x + 1$

Then $\frac{\mathrm{du}}{\mathrm{dx}} = 3 {x}^{2} - x - 3$

~~~~~~~~~~~~~Out one stage ~~~~~~~~~~~~~~~~~~~~

Let $v = {u}^{\frac{2}{5}}$

then $\frac{\mathrm{dv}}{\mathrm{du}} = \frac{2}{5} {u}^{- \frac{3}{5}}$

~~~~~~~~~~~~~~Out another stage ~~~~~~~~~~~~~~~~

Let $y = \ln \left(v\right)$

then $\frac{\mathrm{dy}}{\mathrm{dv}} = \frac{1}{v}$

~~~~~~~~~~~~~~~Putting it all togther~~~~~~~~~~~~~~~~~~~~~

But $\text{ } \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dv}} \times \frac{\mathrm{dv}}{\mathrm{du}} \times \frac{\mathrm{du}}{\mathrm{dx}}$

When you cancel out the right hand side you do indeed end up with $\frac{\mathrm{dy}}{\mathrm{dx}}$ . So the chain rule is just multiplying out the values obtained at each stage