|
(a) One way of defining sec-1 x is to say that y = sec-1x ↔ sec y = x and 0 < y < π/2 or π < y < 3π/2. Show that, with this definition, d/dx (sec-1 x) = 1 / x√x2 – 1
(b) Another way of defining sec-1 that is sometimes used is to say that y = sec-1x ↔ sec y = x and 0 < y < π y ≠ 0. Show that, with this definition, d/dx (sec-1 x) = 1 / x√x2 – 1 |
| New search. (Also 1294 free access solutions) |