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# Find the derivative of f(x)

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

### 2 Answers by Expert Tutors

David T. | Experienced Math and Physics TutorExperienced Math and Physics Tutor
4.8 4.8 (12 lesson ratings) (12)
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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) = x2-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.
Lisa O. | Pre-algebra, Algebra, Geometry, Trig, Pre-calc, CalculusPre-algebra, Algebra, Geometry, Trig, Pr...
5.0 5.0 (154 lesson ratings) (154)
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If the original problem is f(x) = x2/3(x4/3-x1/3+x-5/3), you could solve this problem two ways: use the product rule, or distribute x2/3 and then find the derivative.

Let's first distribute and then find the derivative.
f(x) = x2/3(x4/3-x1/3+x-5/3)
= x2/3+4/3-x2/3+1/3+x2/3+-5/3
= x6/3-x3/3+x-3/3
= x2-x+x-1

f'(x) = 2x2-1-x1-1+(-1)x-1-1
= 2x-x0-x-2
= 2x-1-x-2

Or use the product rule:
f'(x) = (x2/3)*[(4/3)x4/3-1-(1/3)x1/3-1+(-5/3)x-5/3-1]+(x4/3-x1/3+x-5/3)*[(2/3)x2/3-1]
= (x2/3)*[(4/3)x1/3-(1/3)x-2/3+(-5/3)x-8/3]+(x4/3-x1/3+x-5/3)*[(2/3)x-1/3]
= (4/3)x2/3+1/3-(1/3)x2/3+-2/3+(-5/3)x2/3+-8/3+(2/3)x4/3+-1/3-(2/3)x1/3+-1/3+(2/3)x-5/3+-1/3
= (4/3)x3/3-(1/3)x0+(-5/3)x-6/3+(2/3)x3/3-(2/3)x0+(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