If f(x) = frac{x^2}{x+a}, then f^{prime prime | vedclass

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(C) Given $f(x) = \frac{x^2}{x+a}$.Using the quotient rule, $f^{\prime}(x) = \frac{(x+a)(2x) - x^2(1)}{(x+a)^2} = \frac{2x^2 + 2ax - x^2}{(x+a)^2} = \

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$\frac{1}{4a}$

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(C) Given $f(x) = \frac{x^2}{x+a}$.Using the quotient rule, $f^{\prime}(x) = \frac{(x+a)(2x) - x^2(1)}{(x+a)^2} = \frac{2x^2 + 2ax - x^2}{(x+a)^2} = \frac{x^2 + 2ax}{(x+a)^2}$.Now, find $f^{\prime \prime}(x)$ by differentiating $f^{\prime}(x)$ again:$f^{\prime \prime}(x) = \frac{(x+a)^2(2x + 2a) - (x^2 + 2ax)(2(x+a))}{(x+a)^4}$$f^{\prime \prime}(x) = \frac{(x+a)(2x + 2a) - 2(x^2 + 2ax)}{(x+a)^3} = \frac{2x^2 + 4ax + 2a^2 - 2x^2 - 4ax}{(x+a)^3} = \frac{2a^2}{(x+a)^3}$.Substituting $x = a$:$f^{\prime \prime}(a) = \frac{2a^2}{(a+a)^3} = \frac{2a^2}{(2a)^3} = \frac{2a^2}{8a^3} = \frac{1}{4a}$.

If $f(x) = \frac{x^2}{x+a}$, then $f^{\prime \prime}(a)$ is equal to

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