To find the derivative of arccsc x, let us assume that y = arccsc x. Thus, the derivative of arctan x (or) tan -1x (or) inverse tan x is 1 / (1 + x 2). From this, sec 2y = 1 + tan 2y = 1 + x 2. Differentiating both sides with respect to x,īy one of the trigonometric identities, sec 2y - tan 2y = 1. Then by the definition of inverse tan, tan y = x. To find the derivative of arctan x, let us assume that y = arctan x. Thus, the derivative of arccos x (or) cos -1x (or) inverse cos x is 1/√ 1-x². Differentiating both sides with respect to x,īy one of the trigonometric identities, sin 2y + cos 2y = 1. Then by the definition of inverse cos, cos y = x. To find the derivative of arccos x, let us assume that y = arccos x. Thus, the derivative of arcsin x (or) sin -1x (or) inverse sin x is 1/√ 1-x².
Then by the definition of inverse sine, sin y = x. To find the derivative of arcsin x, let us assume that y = arcsin x.
We use the process of implicit differentiation (which is the process of using the chain rule when the functions are implicitly defined) to derive the inverse trig derivatives.
Thus, it is difficult to memorize them unless we know how these formulas are derived. Rather they include squares and square roots. If we observe them carefully, the derivatives neither include trigonometric functions nor include inverse trigonometric functions. In the previous section, we have already seen the formulas of derivatives of inverse trigonometric functions.
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