Could someone kindly show me the steps involved in solving the following.
y = ln(x/1+x^2)
The answer should come out to dy/dx = (1-x^2)/x(1+x^2), but I don't know exactly how this was achieved.
First remember the property of logs that
$\displaystyle \ln\left( \frac{A}{B} \right)=\ln(A)-\ln(B)$
so we get
$\displaystyle y=\ln(x)-\ln(1+x^2)$
Now using the chain rule we get
$\displaystyle y'=\frac{1}{x}-\frac{2x}{1+x^2}$
Now simplifying we get
$\displaystyle y'=\frac{1+x^2}{x(1+x^2)}-\frac{2x^2}{x(1+x^2)}=\frac{1-x^2}{x(1+x^2)}$