## Calculus 8th Edition

$$f'(x)=\frac{2x+2x^2}{(1+2x)^2}$$ $$f''(x)=\frac{2}{(1+2x)^3}$$
$$f(x)=\frac{x^2}{1+2x}$$ Differentiating with respect to $x$, using quotient's rule, $$f'(x)=\frac{(2x)(1+2x)-x^2(2)}{(1+2x)^2}=\frac{2x+2x^2}{(1+2x)^2}$$ Differentiating again with respect to $x$, using quotient's rule, $$f''(x)=\frac{(2+4x)(1+2x)^2-(2x+2x^2)(2(1+2x)(2))}{(1+2x)^4}$$ $$f''(x)=\frac{2(1+2x)^3-8x(1+x)(1+2x)}{(1+2x)^4}=\frac{2(1+2x)^2-8x(1+x)}{(1+2x)^3}$$ $$f''(x)=\frac{2(1+4x^2+4x)-8x-8x^2}{(1+2x)^3}\implies f''(x)=\frac{2}{(1+2x)^3}.$$