Question

$$\text { State the derivatives of } \sin ^{-1} x, \tan ^{-1} x, \text { and } \sec ^{-1} x$$.

Short answer

Question: Find the derivatives of the inverse trigonometric functions \(\sin^{-1}x\), \(\tan^{-1}x\), and \(\sec^{-1}x\). Answer: The derivative of \(\sin^{-1}x\) is \(\frac{1}{\sqrt{1-x^2}}\), the derivative of \(\tan^{-1}x\) is \(\frac{1}{1+x^2}\), and the derivative of \(\sec^{-1}x\) is \(\frac{1}{x\sqrt{x^2-1}}\).

Step-by-step solution

Derivative of \(\sin^{-1}x\)

Step 1: Let \(y = \sin^{-1}x\). Then, we have \(x = \sin y\). Step 2: Differentiate both sides with respect to \(x\) using implicit differentiation. $$\frac{d}{dx}(x) = \frac{d}{dx}(\sin y)$$ $$1 = (\cos y)\frac{dy}{dx}$$ Step 3: Solve for \(\frac{dy}{dx}\): $$\frac{dy}{dx}=\frac{1}{\cos y}$$ Step 4: Use the Pythagorean identity \(\sin^2 y + \cos^2 y = 1\) to express \(\cos y\) in terms of \(x\): $$\cos^2 y = 1 - \sin^2 y= 1 - x^2$$ Step 5: Take the square root of \(\cos^2 y\) and substitute it in the expression for \(\frac{dy}{dx}\): $$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$$ Therefore, the derivative of \(\sin^{-1}x\) is \(\frac{1}{\sqrt{1-x^2}}\).

Derivative of \(\tan^{-1}x\)

Step 1: Let \(y = \tan^{-1}x\). Then, we have \(x = \tan y\). Step 2: Differentiate both sides with respect to \(x\) using implicit differentiation. $$\frac{d}{dx}(x) = \frac{d}{dx}(\tan y)$$ $$1 = (\sec^2 y)\frac{dy}{dx}$$ Step 3: Solve for \(\frac{dy}{dx}\): $$\frac{dy}{dx}=\frac{1}{\sec^2 y}$$ Step 4: Use the identity \(\sec^2 y = 1 + \tan^2 y\) to express \(\sec^2 y\) in terms of \(x\): $$\sec^2 y = 1 + x^2$$ Step 5: Substitute this expression in the expression for \(\frac{dy}{dx}\): $$\frac{dy}{dx}=\frac{1}{1+x^2}$$ Therefore, the derivative of \(\tan^{-1}x\) is \(\frac{1}{1+x^2}\).

Derivative of \(\sec^{-1}x\)

Step 1: Let \(y = \sec^{-1}x\). Then, we have \(x = \sec y\). Step 2: Differentiate both sides with respect to \(x\) using implicit differentiation. $$\frac{d}{dx}(x) = \frac{d}{dx}(\sec y)$$ $$1 = (\sec y \tan y)\frac{dy}{dx}$$ Step 3: Solve for \(\frac{dy}{dx}\): $$\frac{dy}{dx}=\frac{1}{\sec y \tan y}$$ Step 4: Use the identity \(\sec^2 y = 1 + \tan^2 y\) to express \(\tan y\) in terms of \(x\): $$\tan y = \sqrt{\sec^2 y - 1} = \sqrt{x^2 - 1}$$ Step 5: Substitute this expression and \(x = \sec y\) in the expression for \(\frac{dy}{dx}\): $$\frac{dy}{dx}=\frac{1}{x\sqrt{x^2-1}}$$ Therefore, the derivative of \(\sec^{-1}x\) is \(\frac{1}{x\sqrt{x^2-1}}\).

Key concepts

Implicit Differentiation

Implicit differentiation is a technique used when you have an equation involving two variables for which you need to find the derivative. It is extremely useful when dealing with inverse trigonometric functions, as these often imply relationships that aren't easy to isolate. For instance, if you have a function given as \(y = \sin^{-1}x\), it suggests that \(x = \sin y\). Here, \(y\) implicitly depends on \(x\), and we need to differentiate both sides with respect to \(x\). This is where implicit differentiation comes into play.

• First, differentiate both sides of the equation with respect to \(x\). Treat \(y\) as a function of \(x\), meaning wherever you differentiate \(y\), add \( \frac{dy}{dx} \) multiplied by the derivative of that function.
• For example, differentiating \(\sin y\) gives \( \cos y \cdot \frac{dy}{dx} \).
• Then, solve for \( \frac{dy}{dx} \), which will give you the derivative of the inverse trigonometric function.

This method helps in finding derivatives where directly applying differentiation is either difficult or not possible.

Pythagorean Identities

Pythagorean identities are essential trigonometric identities that relate the squares of sine, cosine, and tangent functions. These identities are particularly helpful in calculus, especially when working with inverse trigonometric functions and their derivatives. For example, the basic identity \( \sin^2 y + \cos^2 y = 1\) is frequently used.

• When working to find the derivative of \( \sin^{-1}x\), after implicit differentiation, one might arrive at an expression involving \( \cos y\).
• Because the original equation is \(x = \sin y\), which implies \( \sin y = x\), you can substitute \(x\) for \(\sin y\) and rearrange the Pythagorean identity to find \( \cos y\).
• This often results in \( \cos y = \sqrt{1-x^2} \), which can then be substituted back into the derivative equation.

Using these identities carefully allows us to simplify expressions and solve for derivatives that would otherwise seem complex.

Calculus Derivatives

Calculus derivatives provide us with the necessary tools to understand and interpret the rate at which functions change. When dealing with inverse trigonometric functions, knowing how to find derivatives becomes extremely helpful. Each inverse function such as \( \sin^{-1}x \), \( \tan^{-1}x \), and \( \sec^{-1}x \) has its own specific derivative formula that can be derived through implicit differentiation and the use of trigonometric identities.

• The derivative of \( \sin^{-1}x \) is \( \frac{1}{\sqrt{1-x^2}} \).
• The derivative of \( \tan^{-1}x \) is \( \frac{1}{1 + x^2} \).
• The derivative of \( \sec^{-1}x \) is \( \frac{1}{x \sqrt{x^2-1}} \).

Each derivative has a unique form determined by manipulating related trigonometric identities and simplifying using implicit differentiation. Familiarity with these derivatives allows deeper insights into the behavior of inverse trigonometric functions.