Derivatives of Inverse Trigonometric Functions
This article is part of the MathHelp Tutoring Wiki |
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
sin^{2}(y) + cos^{2}(y) = 1 and from the above the cosine can be substituted with x
sin^{2}(y) + x^{2} = 1 so sin^{2}(y) = 1 - x^{2} and therefore
dy/dx = -1 / sqrt(1 - x^{2}) = f'(x)