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

1 Answer
Tony B
Oct 30, 2015

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

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