Differentiation of inverse trigonometric and hyperbolic functions

Let u

Then du dx

= f ′(x) and dy du

= 1√ 1−u2 (see para. (i))

Thus dy dx

= dy du

× du dx

= 1√ 1− u2 f

= f ′(x)√

1−[ f (x)]2 (v) The differential coefficients of the remaining

inverse trigonometric functions are obtained in a similar manner to that shown above and a summary of the results is shown in Table 59.1.