^{2}-x+x

^{-1},

Find the derivative of f(x)= x^2/3(x^4/3-x^1/3+x^-5/3)

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Assuming Lisa's interpretation of the statement of the problem is correct, this is a problem that was intended to demonstrate the value of remembering your algebraic manipulations while evaluating calculus expressions.

The product rule route creates a mess, as Lisa demonstrated with exacting attention to detail in her second evaluation.

By distributing, we see that a seemingly difficult expression becomes

f(x) = x^{2}-x+x^{-1},

quite quickly, which is a much easier expression to take a derivative of (mostly due to reduced opportunity to make an error). Of course the answers agree as they should.

If the original problem is f(x) = x^{2/3}(x^{4/3}-x^{1/3}+x^{-5/3}), you could solve this problem two ways: use the product rule, or distribute x^{2/3} and then find the derivative.

Let's first distribute and then find the derivative.

f(x) = x^{2/3}(x^{4/3}-x^{1/3}+x^{-5/3})

= x^{2/3+4/3}-x^{2/3+1/3}+x^{2/3+-5/3}

= x^{6/3}-x^{3/3}+x^{-3/3}

= x^{2}-x+x^{-1}

f'(x) = 2x^{2-1}-x^{1-1}+(-1)x^{-1-1}

= 2x-x^{0}-x^{-2}

= 2x-1-x^{-2}

Or use the product rule:

f'(x) = (x^{2/3})*[(4/3)x^{4/3-1}-(1/3)x^{1/3-1}+(-5/3)x^{-5/3-1}]+(x^{4/3}-x^{1/3}+x^{-5/3})*[(2/3)x^{2/3-1}]

= (x^{2/3})*[(4/3)x^{1/3}-(1/3)x^{-2/3}+(-5/3)x^{-8/3}]+(x^{4/3}-x^{1/3}+x^{-5/3})*[(2/3)x^{-1/3}]

= (4/3)x^{2/3+1/3}-(1/3)x^{2/3+-2/3}+(-5/3)x^{2/3+-8/3}+(2/3)x^{4/3+-1/3}-(2/3)x^{1/3+-1/3}+(2/3)x^{-5/3+-1/3}

= (4/3)x^{3/3}-(1/3)x^{0}+(-5/3)x^{-6/3}+(2/3)x^{3/3}-(2/3)x^{0}+(2/3)x^{-6/3}

= (4/3)x-(1/3)+(-5/3)x^{-2}+(2/3)x-(2/3)+(2/3)x^{-2}

= (4/3+2/3)x-(1/3+2/3)+(-5/3+2/3)x^{-2}

= (6/3)x-(3/3)+(-3/3)x^{-2}

= 2x-1-x^{-2}

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