How do you differentiate (x^2-5x+2)/root3xx2−5x+23√x?

1 Answer
Lovecraft
Oct 14, 2015

dy/dx = (5x^(2/3) + 10x^(-5/6) + 2x^(-4/3))/3dydx=5x23+10x−56+2x−433

Explanation:

The simplest way is to put the function into exponential notation, foil them and then derivate

y = (x^2 - 5x + 2)/root(3)(x)y=x2−5x+23√x

y = (x^2 - 5x + 2)*(x^(-1/3))y=(x2−5x+2)⋅(x−13)

y = x^(2-1/3) - 5x^(1-1/3) + 2x^(-1/3)y=x2−13−5x1−13+2x−13

y = x^(5/3) - 5x^(-2/3) + 2x^(-1/3)y=x53−5x−23+2x−13

dy/dx = (5x^(2/3))/3 -(5*(-2)*x^(-5/6))/3 + (2x^(-4/3))/3dydx=5x233−5⋅(−2)⋅x−563+2x−433

dy/dx = (5x^(2/3) + 10x^(-5/6) + 2x^(-4/3))/3dydx=5x23+10x−56+2x−433

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