Example 1: y = tan^(-1)((sqrt(1+$x^2$) - 1)/x) Substitute x = tan t: sqrt(1+$tan^2$ t) = sec t. Argument = $\frac{sec t - 1}{tan}$ t = $\frac{1 - cos t}{(sin t)}$ = tan$\frac{t}{2}$. So y = tan^(-1)(tan$\frac{t}{2}$) = t/2 = $\frac{1}{2}$*tan^(-1)(x). dy/dx = $\frac{1}{2(1+x^2}$).

Example 2: y = sin^(-1)((2^(x+1))/(1+4^x)) Let 2^x = t. Then 2^(x+1) = 2t and 4^x = $t^2$. y = sin^(-1)(2$\frac{t}{1+t^2}$). Put t = tan(u): 2$\frac{t}{1+t^2}$ = sin(2u). y = sin^(-1)(sin(2u)) = 2u = 2*tan^(-1)(2^x) [if |2u| <= pi/2]. dy/dx = 2 * $\frac{1}{1+4^x}$ * 2^x * ln 2 = (2^(x+1) * ln 2)/(1+4^x).

Example 3: y = cos^(-1)$\frac{(1-x^2}{(1+x^2)}$) Put x = tan t: $\frac{1-tan^2 t}{(1+tan^2 t)}$ = cos(2t). y = cos^(-1)(cos(2t)) = 2t = 2*tan^(-1)(x) [for x >= 0]. dy/dx = $\frac{2}{1+x^2}$.

The pattern: inverse trig of a recognizable expression -> simplify -> trivial derivative.