Here's the question you clicked on:
Cake4u
Find F` (x) of f(x)= 1 / 1 + x^2 then calculate f`` (x)
f'(x)= -2x(1+x^2)^-2 f''(x)= -2(1+x^2)^-2 + -2x*-2(1+x^2)^-3 = -2(1+x^2)^-2+4(1+x^2)^-3
\[f(x)=\frac{1}{1+x^2}\] is this ur function?
this chick need to use parenthesis at some point.
\[f[x]=\frac{1}{1+x^2} \]\[f'[x]=-\frac{2 x}{\left(1+x^2\right)^2} \]\[f\text{''}[x]=\frac{8 x^2}{\left(1+x^2\right)^3}-\frac{2}{\left(1+x^2\right)^2} \]\[f\text{'''}[x]=-\frac{48 x^3}{\left(1+x^2\right)^4}+\frac{24 x}{\left(1+x^2\right)^3} \]