Question

求y=arctan(1-x^2)/(1+x)的微分

Answer 1

dy=[darctan(1-x²)*(1+x)-arctan(1-x²)d(1+x)]/(1+x)²
={1/[1+(1-x²)²]d(1-x²)*(1+x)-arctan(1-x²)dx}/(1+x)²
={-2x(1+x)/[1+(1-x²)²]dx-arctan(1-x²)dx}/(1+x)²
={-2x(1+x)-[1+(1-x²)²]*arctan(1-x²)}/{(1+x)²[1+(1-x²)²]} dx

Answer 2

y=f(x)=lg(x+√(x^2+1)
x+√(x^2+1)=10^y
10^y-x=√(x^2+1)
平方
10^2y-2x*10^y+x^2=x^2+1
10^2y-2x*10^y=1
x=(10^2y-1)/2*10^y
=1/2*(10^2y/10^y-1/10^y)
=1/2*(10^y-10^-y)