Like many functions, the trigonometric functions also have inverse. Just like how we can find the derivative of trigonometric functions, we can also find the derivatives of inverse trigonometric functions. In this article we shall take a quick look at inverse trig functions derivatives. To be able to find the standard formulae for derivatives of inverse trig functions, we would need the formula for derivative of any general inverse function.
That would be like this:
f’(x) = 1/g’(f(x)), where f and g are the inverse functions of each other.
Let us first try finding the inverse trig function derivatives of the sine function. The sine function inverse is written as arc sin (x). Therefore the function would look like this: y = arcsin(x),
Therefore, x= sin (y) for –pi/2 ≤ y ≤ pi/2. We fix this domain for the sine function to ensure that our inverse exists. If we don’t restrict the domain, then y could have multiple values for same value of x. For example, sin (pi/4) = 1/sqrt(2) and sin (3pi/4) is also 1/sqrt(2). With that in mind, we can write the relationship between sin and arc sin as follows:
Sin(arc sin (x)) = x and arc sin(sin(x)) = x
Thus here our f(x) = arc sin(x) and g(x) = sin (x), then using the derivative of inverse formula that we stated above we have:
f’(x) = 1/f’(g(x)) = 1/cos (arc sin (x))
This formula may to be applicable in practice. So let us make a few changes in there. We had earlier x = sin y so y = arc sin (x). Using that, the denominator of our derivative would become
Cos(arc sin (x)) = cos y
Next we use our primary trigonometric identity which was:
Cos^2 (x) + sin^2 (x) = 1
Thus, Cos^2 (y) + sin^2 (y) = 1 from this we have
Cos (y) =√( 1- sin^2 (y)) = √(1 – (sin y)^2). But we have already established that sin y = x. So replacing that we get,
Cos (y) = √(1 – x^2). So now plugging all that back to our equation of derivative of inverse trig functions for sine f’(x), we have:
f’(x) = 1/cos(arc sin(x)) = 1/cos y = 1/√(1-x^2)
Therefore we see that derivative of arc sin (x) = 1/√(1-x^2).
The derivatives of other trigonometric inverse functions of arc cos, arc tan etc can be derived in a similar way. They are:
d/dx arc cos (x) = -1/√(1-x^2) and d/dx arc tan (x) = 1/(1+x^2)
