Post a New Question

calculus

posted by .

Hey I am working on differentiation/quotient rule for this problem...can someone help me?

x^2-3/x-2.

  • calculus -

    if y=u/v,
    y' = (u'v-uv')/v^2

    so, let u = x^2-3, u'=2x
    let v = x-2, v'=1

    y' = ((2x)(x-2)-(x^2-3)(1))/(x-2)^2
    = (2x^2-2x-x^2+3)/(x-2)^2
    = (x^2-4x+3)/(x-2)^2
    = (x-3)(x-1)/(x-2)^2

    or, you can always think of it as a product rule:

    y = (x^2-3) * (x-2)^-1

    y' = (2x-3)/(x-2) - (x^2-3)*(x-2)^-2
    which works out the same

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question