Problem 13 - Essential Calculus Early Transcendentals

Question

Prove that \(\frac{d}{d x}\left(\cot ^{-1} x\right)=-\frac{1}{1+x^{2}}\)

Step-by-Step Solution

Answer

The derivative of \( \cot^{-1}(x) \) is \(-\frac{1}{1 + x^2}\).

Step 1: Define Inverse Function

Let \( y = \cot^{-1}(x) \). This implies that \( \cot(y) = x \). We need to differentiate \( \cot^{-1}(x) \) with respect to \( x \).

Step 2: Differentiate Implicitly

Differentiate both sides of \( \cot(y) = x \) with respect to \( x \). Using the derivative of \( \cot(y) \), we have \(-\csc^2(y) \frac{dy}{dx} = 1\).

Step 3: Solve for \( \frac{dy}{dx} \)

Rearrange the equation to solve for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = -\csc^2(y). \]

Step 4: Use Trigonometric Identity

Recall that \( \csc^2(y) = 1 + \cot^2(y) \). Substitute \( \cot(y) = x \) into the identity to get \( \csc^2(y) = 1 + x^2 \).

Step 5: Simplify the Derivative

Substitute \( \csc^2(y) = 1 + x^2 \) back into the expression for \( \frac{dy}{dx} \). This gives:\[ \frac{dy}{dx} = -\frac{1}{1 + x^2}. \]

Step 6: Conclusion

Thus, we have shown that the derivative of \( \cot^{-1}(x) \) is \(-\frac{1}{1 + x^2}\).

Key Concepts

Differentiation

Differentiation is a core concept in calculus that involves finding a function's derivative. The derivative represents how a function changes as its input changes. In other words, it is the rate at which the function varies. This is analogous to measuring the speed of a moving vehicle, where the function is the path, and the derivative is the speed.

• In mathematical terms, the derivative of a function \( f(x) \) is denoted as \( f'(x) \) or \( \frac{df}{dx} \), showing the change in \( f \) with respect to \( x \).
• Derivatives help us understand properties like the rate of change and slope of curves.

For trigonometric functions, differentiation aids in calculating angles and lengths in various settings, such as optimizing shapes or analyzing waveforms. The derivative of an inverse trigonometric function, like \( \cot^{-1}(x) \), allows us to explore how the angle changes as the cotangent itself varies.

Implicit Differentiation

Implicit differentiation is a technique used when dealing with functions that are not given explicitly as \( y = f(x) \), but rather in a form where \( y \) is mixed with \( x \), such as \( \cot(y) = x \).

• It involves differentiating both sides of the equation with respect to \( x \) while treating \( y \) as a function of \( x \).
• This method requires using the chain rule, which states that the derivative of a composite function \( g(f(x)) \) is \( g'(f(x)) \times f'(x) \).

By applying implicit differentiation in our problem, we differentiated \( \cot(y) = x \) to find the expression \( -\csc^2(y) \frac{dy}{dx} = 1 \). Here, the chain rule helps in isolating \( \frac{dy}{dx} \), which is our target derivative.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold for all values of the involved angles. They are crucial tools in simplifying expressions and solving equations.

• Key identities include Pythagorean identities like \( \sin^2(\theta) + \cos^2(\theta) = 1 \) and secant cosecant identities such as \( \csc^2(\theta) = 1 + \cot^2(\theta) \).
• These identities allow transformations and substitutions that simplify calculus tasks.

In our exercise, the identity \( \csc^2(y) = 1 + \cot^2(y) \) was pivotal. We substituted \( \cot(y) = x \) to express \( \csc^2(y) \) in terms of \( x \), simplifying our derivative expression to be \( \frac{dy}{dx} = -\frac{1}{1 + x^2} \). Thus, trigonometric identities bridge gaps between concepts, facilitating complex calculus operations.