Question

найти вторую производную от ln(x^2+1)

Answer 1

$f(x)=ln(x^2+1)\\f'(x)=\frac{2x}{x^2+1}\\f''(x)=\frac{(2x)'(x^2+1)-2x(x^2+1)'}{(x^2+1)^2}\\f''(x)=\frac{2(x^2+1)-4x^2}{(x^2+1)^2} \\f''(x)=\frac{2-2x^2}{(x^2+1)^2}$

Answer 2

y'=(ln(x²+1))'=2x/(x²+1)

y''=(2x/(x²+1))'=(2*(x²+1)-2x*2x)/((x²+1)²)=(2-2x²)/((x²+1)²)