Inverse Functions: Solutions

y = f(x) = tan^-1 x
x = f^-1(y) = tan y
$dy dx &sp;=&sp; 1 D_y_(tan &sp;y)$ $&sp;=&sp; 1 sec^2^y &sp;=&sp; 1 1 + tan^2^y &sp;=&sp; 1 1 + x^2^$
This result is also derived implicitly in Derivatives of Inverse Trig Functions.
y = x^1/3
x = f^-1(y) = y³
$dy dx &sp;=&sp; 1 D_y_y^3^$ $&sp;=&sp; 1 3y^2^&sp;=&sp; 1 3x^2/3^&sp;=&sp; 13 &sp; x^-2/3^$
This agrees with the power rule applied to fractional exponents.

jjnichol@mit.edu