Derivative?

Question

f(x)= [( x^2 + 6 x + 8 )/( 3 x + 12 )]

Find f'(5)
and f''(5)

Answer 1

Simplify first!

f(x) = [(x + 2) (x + 4)] / [3 (x+4)]
= (x+2) / 3, for x ≠ -4.
So f'(x) = 1/3 and f''(x) = 0 (and hence f'(5) = 1/3, f''(5) = 0).

If you try to tackle this directly with the quotient rule, it gets a bit messy!

f'(x) = [(2x + 6) (3x + 12) - (x^2 + 6x + 8) (3)] / [(3x + 12)^2]
= [6x^2 + 42x + 72 - 3x^2 - 18x - 24] / [(3x + 12)^2]
= (3x^2 + 24x + 48) / [(3x + 12)^2]
Of course we can simplify this to
= [3 (x^2 + 8x + 16)] / [9 (x + 4)^2]
= (x + 4)^2 / [3 (x+4)^2]
= 1/3.

I won't bother trying to differentiate the unsimplified version of f'(x). You can try it as an exercise if you like!

Answer 2

let u=(x^2+6x+8)
v=1 / (3x+12)
derivative = udv+vdu
(x^2+6x+8) (-1/(3x+12)^2)(3) + [1/(3x+12)][2x+6)
=-3(x^2+6x+8)/(3x+12)^2 +(2x+6)/(3x+12)