How to find the second derivative of $f(x)=\frac{x^2}{10+x}$ How do you find the second derivative of $f(x)=\frac{x^2}{10+x}$?
I get $f'(x)={20x+x^2}/{(10+x)^2}$
Then get stuck here:
$(10+x)^2 (20+2x) - (20x+x^2) \frac {2(10+x)}{((10+x)^2)^2}$

f(x) = $\frac{x^2}{10+x}$
You need to use the quotient rule: $(\frac{f}{g})'=\frac{f'*g-g'*f}{g^2}$
Then f= $x^2$ take the derivative of that and you get $\rightarrow f'=2x$
$g= 10+x$, take the derivative of that and you get $\rightarrow g'= 1$
Now let us apply the quotient rule and use our following notations. 
$f'(x)= \frac{(2x)*10+x - ((1)*x^2)}{(10+x)^2}$
Now let us simplify and we get $\rightarrow \frac{20x+x^2}{(10+x)^2}$
Let us get $f''(x)$, but first let us get the notations ready to substitute them.
$f'= 20x+x^2 \rightarrow f''=20+2x$  & $g'= (10+x)^2 \rightarrow g''= 2(10+x)$
Then we use the quotient rule again and we get:
$f''(x)= \frac{(2x+10)*(10+x)^2 - (20x+x^2(2(10+x))}{((10+x)^2)^2}$
We simplify and get $\rightarrow f''(x)= \frac{200}{(x+10)^3}$
