<s>[INST] 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:
$(10+x)^2 (20+2x) - (20x+x^2) \frac {2(10+x)}{((10+x)^2)^2}$
 [/INST] Just do it:
$f(x) = \frac {x^2}{10 - x}$
$f'(x) = \frac {2x(10 - x) - x^2(-1)}{(10-x)^2} = \frac{-2x^2 + 20x +x^2}{(10-x)^2} = \frac{20x - x^2}{(10 - x)^2}=\frac {x(20-x)}{(10-x)^2}$
$f''(x) =\frac{[x(-1)+ (20-x)](10-x)^2 - x(20 -x)2(10-x)(-1)}{(10-x)^4}$
$=\frac{[x(-1)+ (20-x)](10-x)^{\not 2} - x(20 -x)2\not{(10-x)}(-1)}{(10-x)^{\not 43}}$
$=\frac{[-x+ (20-x)](10-x) + 2x(20 -x)}{(10-x)^3}$
$=\frac{2(10-x)^2 + 2x(20-x)}{(10-x)^3}=\frac{200 - 40x + 2x^2 + 40x - 2x^2}{(10-x)^3}=\frac{200}{(10-x)^3}$
Or you can generalize first:
$(\frac fg)'' =$
$((\frac fg)')' = $
$(\frac{f'g - fg'}{g^2})'=$
$(\frac{f'}g)' - (\frac{fg'}{g^2}) =$
$\frac{f''g - f'g'}{g^2} - \frac{(f'g'+fg'')g^2 - fg'*2gg'}{g^4} =$
$\frac{f''}g - \frac{f'g'}{g^2} - \frac{f'g'}{g^2} - \frac{fg''}{g^2} + \frac{2fg'^2}{g^3}=$
$\frac{f''}g - 2\frac{f'g'}{g^2} - \frac{fg''}{g^2} + \frac{2fg'^2}{g^3}=$
$\frac{2}{10-x} - 2\frac{-2x}{(10-x)^2} - 0 + \frac{2x^2}{(10-x)^3}=$
$\frac {2(10-x)^2 + 2x(10 - x) + 2x^2}{(10-x)^3}=$
$\frac{2[(10-x)+x]^2}{(10-x)^3 }=\frac {2*10^2}{(10-x)^3}= \frac {200}{(10-x)^2}$
But I can't see that $(\frac fg)'' = \frac{f''}g - 2\frac{f'g'}{g^2} - \frac{fg''}{g^2} + \frac{2fg'^2}{g^3}$ is anything worth memorizing.... but still....
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