However dy dx
= 1dx dy
Hence, when y = sin−1 x then dy dx
= 1√ 1 − x2
(ii) A sketch of part of the curve of y = sin−1 x is shown in Fig. 33(a). The principal value of sin−1 x is defined as the value lying between −π/2 and π/2. The gradient of the curve between points A and B is positive for all values of x and thus only the positive value is taken when evaluating
1√ 1 − x2 .