Answer:
27/64 or any other equivalent form
Step-by-step explanation:
using quotient rule
[tex]\frac{d}{dx} (\frac{f}{g} )= \frac{f'(g)-g'(f) }{g^2}[/tex]
f'(x) = [tex]\frac{2x(3+x)-x^2}{(3+x)^2}[/tex] = [tex]\frac{6x+2x^2 -x^2}{(3+x)^2} = \frac{x^2+6x}{(3+x)^2}[/tex]
using quotient rule again
f''(x) = [tex]\frac{(2x+6)(3+x)^2 - 2(3+x)(x^2 + 6x)}{(3+x)^4}[/tex]
simplifying to make it easier to plug in
f''(x) = [tex]\frac{(2x+6)(x^2+6x+9)- (6+2x)(x^2+6x)}{(x+3)^4}[/tex]
= [tex]\frac{(2x^3+12x^2+18x+6x^2+36x+9)-(6x^2+36x+2x^3+12x^2)}{(x+3)^4}[/tex]
= [tex]\frac{2x^3+12x^2+18x+6x^2+36x+9-6x^2-36x-2x^3-12x^2}{(x+3)^4}[/tex]
= [tex]\frac{18x+9}{(x+3)^4}[/tex]
f''(1) = [tex]\frac{18+9}{4^4}[/tex] = [tex]\frac{27}{64}[/tex]
