Find the derivative of f(x)

Question

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

Answer 1

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.

Answer 2

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

= x²-x+x^-1

 

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

= 2x-x⁰-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)x⁰+(-5/3)x^-6/3+(2/3)x^3/3-(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