derivative question....?

Question

how do i find the derivative of f(x) = (x^3 + x^2 + x) / (x^4 + 3x^2 + 1)

Answer 1

Quotient rule:

f'(x) = [(x^4 + 3x^2 + 1)*d/dx(x^3 + x^2 + x) - (x^3 + x^2 + x)d/dx(x^4 + 3x^2 + 1)] / (x^4 + 3x^2 + 1)^2
= [(x^4 + 3x^2 + 1)(3x^2 + 2x) - (x^3 + x^2 + x)(4x^3 + 6x)] / (x^4 + 3x^2 + 1)^2

Then simplify from there.

Answer 2

Yep, quotient rule -- bottom times deriv of top minus top times deriv of bottom, all over the bottom squared. (That's how I always remembered it at test time!)

That looks like it will get messy. You might try to see if you can factor something out of both the numerator and denominator before you tackle the problem. Sometimes you can, and that makes it a little easier to deal with.

Answer 3

The quotient rule says:

d(f(x)/g(x) = {g(x)f'(x) - f(x)g'(x)}g^2(x)

"Bottom times derivative of top minus top times derivative of bottom all divided by the bottom squared."