If \( f(x)=\frac{x^{2}}{3+x} \), find \( f^{\prime \prime}(3) \) \( f^{\prime \prime}(3)=\square \)

To find \( f^{\prime \prime}(3) \), we first need to compute the first and second derivatives of \( f(x) \). 1. **First derivative \( f'(x) \)**: Using the quotient rule, where \( f(x) = \frac{u}{v} \) with \( u = x^2 \) and \( v = 3 + x \): \[ f'(x) = \frac{u'v - uv'}{v^2} = \frac{(2x)(3+x) - (x^2)(1)}{(3+x)^2} \] Simplifying: \[ = \frac{2x(3+x) - x^2}{(3+x)^2} = \frac{6x + 2x^2 - x^2}{(3+x)^2} = \frac{x^2 + 6x}{(3+x)^2} \] 2. **Second derivative \( f''(x) \)**: We apply the quotient rule again to \( f'(x) \): Let \( u = x^2 + 6x \) and \( v = (3+x)^2 \). Then, we need \( u' \) and \( v' \): \[ u' = 2x + 6 \] \[ v = (3+x)^2 \Rightarrow v' = 2(3+x) \] Now applying the quotient rule: \[ f''(x) = \frac{u'v - uv'}{v^2} \] This becomes: \[ = \frac{(2x+6)(3+x)^2 - (x^2+6x)(2(3+x))}{(3+x)^4} \] 3. **Evaluating \( f''(3) \)**: First we substitute \( x = 3 \): - \( u = 3^2 + 6(3) = 9 + 18 = 27 \) - \( v = (3+3)^2 = 6^2 = 36 \) - \( u' = 2(3) + 6 = 6 + 6 = 12 \) - \( v' = 2(6) = 12 \) Now, substituting these values into the second derivative: \[ f''(3) = \frac{(12)(36) - (27)(12)}{36^2} \] This simplifies to: \[ = \frac{432 - 324}{1296} = \frac{108}{1296} = \frac{1}{12} \] So, we find that \( f^{\prime \prime}(3) = \frac{1}{12} \). Answer: \( f^{\prime \prime}(3) = \frac{1}{12} \)