Higher order derivatives...?

Question

How would you find the first and second derivatives of this function:
(x+3)/(x^2 +2x)? x plus 3 over x squared plus 2x is the way this should be read in case i wrote it wrong.

Answer 1

Quotient Rule:
let Hi be on top (it's high)
let Ho be on bottom (it's low, which rhymes with Ho)

Your original fraction is Hi / Ho.
The derivative is "Ho d Hi minus Hi d Ho over Ho Ho." That's the way I teach it to my students, anyway... just 'cause it's fun and goofy to say.

first derivative: d/dx [(x + 3) / (x² + 2x)]
= [(x² + 2x) d/dx(x + 3) - (x + 3) d/dx(x² + 2x)] / (x² + 2x)²
= [(x² + 2x) (1) - (x + 3) (2x + 2)] / (x² + 2x)²
= [(x² + 2x) - (2x² +8x + 6)] / (x² + 2x)²
= (-x² - 6x - 6) / (x² + 2x)²

second derivative: d/dx [(-x² - 6x - 6) / (x² + 2x)²]
= [(x² + 2x)² d/dx (-x² - 6x - 6) - (-x² - 6x - 6) d/dx (x² + 2x)²] / (x² + 2x)^4
= [(x² + 2x)² (-2x - 6) - (-x² - 6x - 6) · 2(x² + 2x)¹(2x + 2)] / (x² + 2x)^4
= [(x^4 + 4x³ + 4x²) (-2x - 6) - (-x² - 6x - 6)(2x² + 4x)(2x + 2)] / (x² + 2x)^4
= [(-2x^5 - 8x^4 - 8x³ - 6x^4 - 24x³ - 24x²) + (x² + 6x + 6)(4x³ + 12x² + 8x)] / (x² + 2x)^4
= [(-2x^5 - 8x^4 - 8x³ - 6x^4 - 24x³ - 24x²) + (4x^5 + 12x^4 + 8x³ + 24x^4 + 72x³ + 48x² + 24x³ + 72x² + 48x)] / (x² + 2x)^4
= [2x^5 - 2x^4 + 72x³ + 96x² + 48x] / (x² + 2x)^4
= 2x(x^4 - x³ + 36x² + 48x + 24) / (x² + 2x)^4

...if I did my maths right.

Answer 2

By repeated application of the quotient rule ( which is really just a convenient representation of the product rule for derivatives) It's fairly simple but takes a bit of time. Here's the rule:

http://mathworld.wolfram.com/QuotientRule.html

In your case: f(x) = x+3 and g(x) = (x^2 +2x) = x(x+2)

Answer 3

1st: 1/(x^2 +2x) - (x+3)(2x +2)/(x^2 +2x)^2
2nd : -(2x +2)/(x^2 +2x)^2 - (2x+2 +x+3)/(x^2 +2x)^2
+ 2(x+3)(2x +2)^2/(x^2 +2x)^3

you might want to simplify this

Answer 4

(x + 3)/(x^2 + 2x)

f'(x) = (f'(x)g(x) - f(x)g'(x))/(g(x)^2)

f(x) = (x + 3)
g(x) = (x^2 + 2x)

f'(x) = (1/(x^2 + 2x)) - (((x + 3)(2x + 2))/((x^2 + 2x)^2)))
f"(x) = ((2(x + 3)(2x + 2)^2) - ((2(2x + 2))/((x^2 + 2x)^2)) - ((2(x + 3))/((x^2 + 2x)^2)))