Question

$$\text { State the derivative formulas for } \sin ^{-1} x, \tan ^{-1} x, \text { and } \sec ^{-1} x$$

Short answer

Question: State the derivative formulas for the inverse trigonometric functions: arcsine (sin⁻¹x), arctangent (tan⁻¹x), and arcsecant (sec⁻¹x). Answer: The derivative formulas for the given inverse trigonometric functions are: $$ \frac{d}{dx}\big(\sin^{-1}(x)\big) = \frac{1}{\sqrt{1-x^2}} $$ $$ \frac{d}{dx}\big(\tan^{-1}(x)\big) = \frac{1}{1+x^2} $$ $$ \frac{d}{dx}\big(\sec^{-1}(x)\big) = \frac{1}{x\sqrt{x^2-1}} $$

Step-by-step solution

Derivative of arcsine (sin⁻¹x)

To find the derivative of sin⁻¹x, first let y = sin⁻¹x. That means sin(y) = x. Differentiate both sides with respect to x using implicit differentiation: $$ \frac{d}{dx}(\sin(y)) = \frac{d}{dx}(x) $$ So, cos(y) * (dy/dx) = 1. Now, isolate dy/dx and use the Pythagorean identity for sine and cosine (sin²(y) + cos²(y) = 1) to replace cos(y): $$ \frac{dy}{dx} = \frac{1}{\cos(y)} = \frac{1}{\sqrt{1-\sin^2(y)}} $$ Since sin(y) = x, the derivative formula is: $$ \frac{d}{dx}\big(\sin^{-1}(x)\big) = \frac{1}{\sqrt{1-x^2}} $$

Derivative of arctangent (tan⁻¹x)

To find the derivative of tan⁻¹x, first let y = tan⁻¹x. That means tan(y) = x. Differentiate both sides with respect to x using implicit differentiation: $$ \frac{d}{dx}(\tan(y)) = \frac{d}{dx}(x) $$ So, (sec²(y))*(dy/dx) = 1. Now, isolate dy/dx and use the Pythagorean identity for tangent and secant (tan²(y) + 1 = sec²(y)) to replace sec²(y): $$ \frac{dy}{dx} = \frac{1}{\sec^2(y)} = \frac{1}{1+\tan^2(y)} $$ Since tan(y) = x, the derivative formula is: $$ \frac{d}{dx}\big(\tan^{-1}(x)\big) = \frac{1}{1+x^2} $$

Derivative of arcsecant (sec⁻¹x)

To find the derivative of sec⁻¹x, first let y = sec⁻¹x. That means sec(y) = x. Differentiate both sides with respect to x using implicit differentiation: $$ \frac{d}{dx}(\sec(y)) = \frac{d}{dx}(x) $$ So, (sec(y)*tan(y))*(dy/dx) = 1. Now, isolate dy/dx and use the Pythagorean identity for secant and tangent (sec²(y) - 1 = tan²(y)) to replace tan(y): $$ \frac{dy}{dx} = \frac{1}{\sec(y)\tan(y)} = \frac{1}{x\sqrt{\sec^2(y)-1}} $$ Reusing the Pythagorean identity for secant (sec²(y) - 1 = tan²(y)), we get: $$ \frac{dy}{dx} = \frac{1}{x\sqrt{x^2-1}} $$ So, the derivative formula for arcsecant is: $$ \frac{d}{dx}\big(\sec^{-1}(x)\big) = \frac{1}{x\sqrt{x^2-1}} $$ In summary, the derivative formulas for the given inverse trigonometric functions are: $$ \frac{d}{dx}\big(\sin^{-1}(x)\big) = \frac{1}{\sqrt{1-x^2}} $$ $$ \frac{d}{dx}\big(\tan^{-1}(x)\big) = \frac{1}{1+x^2} $$ $$ \frac{d}{dx}\big(\sec^{-1}(x)\big) = \frac{1}{x\sqrt{x^2-1}} $$

Key concepts

Derivative of Arcsine

Understanding the derivative of arcsine is essential for solving various calculus problems. The arcsine function, denoted as \(\sin^{-1}(x)\), represents the inverse of the sine function. To determine its derivative, we use implicit differentiation, which is a powerful technique for finding the derivative of functions that are implicitly defined.

Starting with \(y = \sin^{-1}(x)\) and the corresponding sine function \(\sin(y) = x\), and differentiating both sides with respect to \(x\), we find that \(\cos(y) \cdot \frac{dy}{dx} = 1\). Applying the Pythagorean identity \(\sin^2(y) + \cos^2(y) = 1\), we can isolate \(\cos(y)\) and solve for \(\frac{dy}{dx}\). This leads us to the derivative formula for arcsine: \(\frac{d}{dx}\big(\sin^{-1}(x)\big) = \frac{1}{\sqrt{1-x^2}}\).

It's important to remember that this derivative only holds when \(x\) is between -1 and 1, since those are the domain limits for the arcsine function.

Derivative of Arctangent

Similar to arcsine, the derivative of arctangent can be found using implicit differentiation. Let's consider \(y = \tan^{-1}(x)\) meaning that \(\tan(y) = x\). Differentiating both sides with respect to \(x\) gives us \(\sec^2(y) \cdot \frac{dy}{dx} = 1\).

By employing the Pythagorean identity for tangent and secant, \(\tan^2(y) + 1 = \sec^2(y)\), we can express \(\frac{dy}{dx}\) in terms of \(x\), resulting in \(\frac{d}{dx}\big(\tan^{-1}(x)\big) = \frac{1}{1+x^2}\).

This derivative reveals how the slope of the tangent line behaves with respect to changes in \(x\). It is valid for all real numbers since arctangent has an infinite range.

Derivative of Arcsecant

The derivative of arcsecant, indicated by \(\sec^{-1}(x)\), may appear more complex, but it can still be tackled using the same strategy of implicit differentiation. After identifying \(y = \sec^{-1}(x)\) and acknowledging the relationship \(\sec(y) = x\), differentiating both sides with respect to \(x\) yields \(\sec(y)\tan(y) \cdot \frac{dy}{dx} = 1\).

Applying the Pythagorean identity \(\sec^2(y) - 1 = \tan^2(y)\) allows us to isolate \(\frac{dy}{dx}\) and obtain the derivative formula for arcsecant: \(\frac{d}{dx}\big(\sec^{-1}(x)\big) = \frac{1}{x\sqrt{x^2-1}}\).

This result is crucial for understanding the rate of change of the arcsecant function, and it holds true for all \(x > 1\) or \(x < -1\), reflecting the domain of the arcsecant function.

Implicit Differentiation

Implicit differentiation is a versatile technique used when dealing with equations defining a function implicitly rather than explicitly. It's a go-to method when functions are given in a form that is difficult or impossible to solve for one variable in terms of another.

In the context of inverse trigonometric functions, this approach is particularly helpful. We differentiate both sides of the equation with respect to \(x\), treating \(y\) as a function of \(x\). After differentiation, we solve for \(\frac{dy}{dx}\) to find the derivative of the original function. Implicit differentiation not only simplifies the calculation of certain derivatives but is also a cornerstone for understanding related rates and solving differential equations.

Pythagorean Identities

Pythagorean identities play a fundamental role in trigonometry and calculus by providing relationships between the trigonometric functions of an angle. These identities are based on the Pythagorean theorem and are used extensively when simplifying expressions and solving equations.

The three main identities include \(\sin^2(\theta) + \cos^2(\theta) = 1\), \(\tan^2(\theta) + 1 = \sec^2(\theta)\), and \(\cot^2(\theta) + 1 = \csc^2(\theta)\). They are especially important in calculus when manipulating the derivatives of inverse trigonometric functions.

With the knowledge of these identities, one can solve complex trigonometric problems by transforming them into a more manageable form, which was exemplified in the derivatives of arcsine, arctangent, and arcsecant calculations.