Expand the following fractions.calculus

expand the following fractions

f1(x)= 6/(x+1)(x+2)(x+3)

f2(x)= 4/(x+1)(x+2)^2

f3(x)= 2/(x+1)(x^2+2x+2)

I would first expand the denominator so I can avoid using the product rule, but I'll leave it as-is in the denominator of the derivative:

f'(x) = 6 / (x + 1)(x² + 5x + 6)
= 6 / x³ + 5x² + 6x + x² + 5x + 6)
= 6 / x³ + 6x² + 11x + 6
Now use quotient rule:
(x³ + 6x² + 11x + 6)(6)' - (6)(x³ + 6x² + 11x + 6)' / ((x+1)(x+2)(x+3))²
= 0 - 6(3x² + 12x + 11) / ((x+1)(x+2)(x+3))²
= -6(3x² + 12x + 11) / ((x+1)(x+2)(x+3))²

f''(x)= 4/(x+1)(x+2)²
f''(x)= 4/(x + 1)(x² + 4x + 4)
f''(x)= 4/(x³ + 4x² + 4x + x² + 4x + 4)
f''(x)= 4/(x³ + 5x² + 8x + 4)
Now use Quotient Rule
(x³ + 5x² + 8x + 4)(4)' - (4)(x³ + 5x² + 8x + 4)' / [(x+1)(x+2)²]²
0 - 4(3x² + 10x + 8) / [(x+1)(x+2)²]²
-4(3x² + 10x + 8) / [(x+1)(x+2)²]²

f'''(x)= 2/(x + 1)(x² + 2x + 2)
f'''(x)= 2/(x³ + 2x² + 2x + x² + 2x + 2)
f'''(x)= 2/(x³ + 3x² + 4x+ 2)
Now use Quotient Rule:
(x³ + 3x² + 4x+ 2)(2)' - (2)(x³ + 3x² + 4x+ 2)' / [(x + 1)(x² + 2x + 2)]²
0 - (2)(3x² + 6x + 4)/ [(x + 1)(x² + 2x + 2)]²
-2(3x² + 6x + 4)/ [(x + 1)²(x² + 2x + 2)²]

Edit: I guess I assumed, since you said this was Calculus, that you'd be looking for the derivative, also. :-)

No problem. The process of expanding the denominators first makes finding the derivative MUCH easier. If you didn't expand first, you'd have to use the Product Rule on the denominator first, which would've been REALLY messy to work with. :-)

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