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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