Aplicando la regla de la cadena a h(x)=sen−1(g(x))h (x) = sen^{− 1} (g (x))h(x)=sen−1(g(x)), tenemos h′(x)=1(1−(g(x))2 g′(x)h'(x) = \frac{1}{(1− (g (x))^2}\,\,g'(x)h′(x)=(1−(g(x))21g′(x)
Ahora sea g(x)=2x3g (x) = 2x^3g(x)=2x3, entonces g′(x)=6x2g'(x) = 6x^2g′(x)=6x2. Sustituyendo en el resultado anterior, obtenemos h′(x)=11−4x6⋅6x2=6x21−4x6h'(x) = \frac{1}{\sqrt{1−4x^6}}⋅6x^2 = \frac{6x^2} {\sqrt{1−4x^6}}h′(x)=1−4x61⋅6x2=1−4x66x2