derive $$\sinh(x+y)=\tanh^{-1}(\frac{x}{y})$$
$$\cosh(x+y)\cdot(1+f')=\frac{\frac{y-xf'}{y^2}}{1-(\frac{x}{y})^2}$$
$$\cosh(x+y)\cdot(1+f')=\frac{\frac{y-xf'}{y^2}}{\frac{y^2-x^2}{y^2}}$$
$$\cosh(x+y)\cdot(1+f')=\frac{y-xf'}{y^2-x^2}$$
$$(y^2-x^2)\cosh(x+y)\cdot(1+f')=y-xf'$$
$$(y^2-x^2)\cosh(x+y)+(y^2-x^2)\cosh(x+y)f'=y-xf'$$
$$((y^2-x^2)\cosh(x+y)+x)f'=y-(y^2-x^2)\cosh(x+y)$$
$$f'=\frac{y-(y^2-x^2)\cosh(x+y)}{(y^2-x^2)\cosh(x+y)+x}$$
is it correct?