arccos(x) derivative | What is derivative of arccosx and cos^-1x?

arccos(x) derivative | What is derivative of arccosx and cos^-1x?

Understanding the Derivative of Arccos(x)

The arccosine function, denoted as ( \text{arccos}(x) ) or ( \cos^{-1}(x) ), is the inverse function of the cosine function restricted to the interval ($0, \pi$). This function is vital in various fields such as trigonometry, calculus, and even in real-world applications such as physics and engineering.

What is the Derivative of Arccos(x)?

To find the derivative of ( y = \text{arccos}(x) ), we start by understanding that ( \text{arccos}(x) ) answers the question: "What angle ( y ) gives me a cosine of ( x )?" By definition, we have:

$x = \cos(y), \quad \text{for } y \in $0, \pi$.$

To find the derivative ( \frac{dy}{dx} ), we can implicitly differentiate both sides of the equation with respect to ( x ). Using the chain rule on the left, we get:

$\frac{dx}{dy} = -\sin(y).$

Now, we also need to express ( \frac{dx}{dy} ) in terms of ( x ) to find ( \frac{dy}{dx} ). By taking the reciprocal, we obtain:

$\frac{dy}{dx} = -\frac{1}{\sin(y)}.$

Now, to express this in terms of ( x ), we can use the identity involving the sine function. Since ( \sin^2(y) + \cos^2(y) = 1 ), we know:

$\sin^2(y) = 1 - x^2.$

Thus, we can rewrite ( \sin(y) ) as:

$\sin(y) = \sqrt{1 - x^2}.$

Substituting this back into our derivative, we get:

$\frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}}.$

Conclusion

In summary, the derivative of the arccosine function is given by:

$\frac{d}{dx} \left( \text{arccos}(x) \right) = -\frac{1}{\sqrt{1 - x^2}} \quad \text{for } x \in (-1, 1).$

This derivative is useful in various applications, particularly when solving problems involving rates of change, optimization, and integration in calculus. Understanding the behavior of the arccos function and its derivative allows for deeper insights into mathematical models involving angles and periodic functions.