Derivatives of Inverse Trigonometric Functions

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It is possible to extract the formula required for the calculation of the derivative of inverse trigonometric functions. The following is an example.

Example: Calculate the derivative of function f given by the following,

y = f(x) = arccos(x)

Solution: We are obviously looking for f'(x)=dy/dx. The above equation is rewritten in the following format,

y = arccos(x)

Based on the definition of "arccos" function, it can be concluded that,

x = cos(y);

We differentiate both sides of the above equation with respect to y (it is noted that the derivative of the trigonometric function "cos" is known and is "-sin")

dx / dy = d(cos(y)) / dy and therefore dx/dy = -sin(y)

It can be therefore concluded that dy/dx = -1/sin(y)

we also know that

sin2(y) + cos2(y) = 1 and from the above the cosine can be substituted with x

sin2(y) + x2 = 1 so sin2(y) = 1 - x2 and therefore

dy/dx = -1 / sqrt(1 - x2) = f'(x)