Question

If f(x) = alt="\frac{x^{2} }{3+x}" align="absmiddle" class="latex-formula"> , find f ''(1).

f ''(1) = ????

Answer 1

Answer:

27/64 or any other equivalent form

Step-by-step explanation:

using quotient rule

$\frac{d}{dx} (\frac{f}{g} )= \frac{f'(g)-g'(f) }{g^2}$

f'(x) = $\frac{2x(3+x)-x^2}{(3+x)^2}$ = $\frac{6x+2x^2 -x^2}{(3+x)^2} = \frac{x^2+6x}{(3+x)^2}$

using quotient rule again

f''(x) = $\frac{(2x+6)(3+x)^2 - 2(3+x)(x^2 + 6x)}{(3+x)^4}$

simplifying to make it easier to plug in

f''(x) = $\frac{(2x+6)(x^2+6x+9)- (6+2x)(x^2+6x)}{(x+3)^4}$

= $\frac{(2x^3+12x^2+18x+6x^2+36x+9)-(6x^2+36x+2x^3+12x^2)}{(x+3)^4}$

= $\frac{2x^3+12x^2+18x+6x^2+36x+9-6x^2-36x-2x^3-12x^2}{(x+3)^4}$

= $\frac{18x+9}{(x+3)^4}$

f''(1) = $\frac{18+9}{4^4}$ = $\frac{27}{64}$