Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find \(D_{x} y\). $$ y=\tanh (\cot x) $$

Step-by-Step Solution

Answer

The derivative is \( D_{x} y = -\text{sech}^2(\cot(x)) \cdot \csc^2(x) \).

Step 1: Apply Chain Rule

The function provided is a composition of two functions: the hyperbolic tangent function \( \tanh(u) \) and the trigonometric function \( u = \cot(x) \). By the chain rule, the derivative of \( y = \tanh(u) \) with respect to \( x \) is given by: \[ D_{x} y = \frac{dy}{du}\cdot\frac{du}{dx} \].

Step 2: Differentiate Outer Function

The derivative of the hyperbolic tangent function is \( \frac{d}{du} \tanh(u) = \text{sech}^2(u) \). Therefore, we have:\[ \frac{dy}{du} = \text{sech}^2(\cot(x)). \]

Step 3: Differentiate Inner Function

The derivative of the cotangent function is \( \frac{d}{dx} \cot(x) = -\csc^2(x) \). Hence, we get:\[ \frac{du}{dx} = -\csc^2(x). \]

Step 4: Apply Chain Rule to Find Derivative

Substitute the results from Steps 2 and 3 into the chain rule expression from Step 1:\[ D_{x} y = \text{sech}^2(\cot(x)) \cdot (-\csc^2(x)). \] Therefore, the derivative \( D_{x} y \) is:\[ D_{x} y = -\text{sech}^2(\cot(x)) \cdot \csc^2(x). \]

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus that allows us to differentiate composite functions. A composite function occurs when one function is applied to the result of another function. For instance, if you have a function like \(y = \tanh(\cot x)\), it combines the hyperbolic function \(\tanh(u)\) where \(u = \cot x\).

The magic of the chain rule lies in its ability to simplify the differentiation of these nested functions. It articulates that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. Simply put:

• If \(y = f(g(x))\), then the derivative \(D_{x}y = f'(g(x)) \cdot g'(x)\).

Understanding this rule is crucial because many real-world problems involve composite functions. Getting comfortable with the chain rule serves to ease the complexity in solving such problems.

Trigonometric Functions

Trigonometric functions play a significant role in calculus, often appearing as the inner functions in composite calculus problems. In our problem, the trigonometric function involved is the cotangent, denoted as \(\cot(x)\).

Trigonometric functions are based on the angles and ratios in a right triangle. Most common trigonometric functions include sine, cosine, and tangent, with cotangent being the reciprocal of tangent:

• \(\cot(x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)}\)

Differentiating \(\cot(x)\) requires using the derivative rules for trigonometric functions:

• The derivative of \(\cot(x)\) is \(-\csc^2(x)\).

This is vital because, as the inner function, its derivative is essential in applying the chain rule for the overall differentiation of the composite function.

Hyperbolic Functions

Hyperbolic functions are similar to trigonometric functions but relate to the geometry of hyperbolas rather than circles. In the exercise, the hyperbolic tangent function \(\tanh(u)\) appears as the outer function.

Hyperbolic functions have direct analogs to the more familiar trigonometric functions. The hyperbolic tangent is often compared to the regular tangent but has properties that make it useful in different contexts, such as modeling real-world phenomena that involve growth or decay:

• \(\tanh(u) = \frac{\sinh(u)}{\cosh(u)}\)

When differentiating \(\tanh(u)\), it’s crucial to remember its derivative:

• \(\frac{d}{du} \tanh(u) = \text{sech}^2(u)\)

This function and its derivative reflect how certain systems change in relation to each other, and knowing how to manipulate them is essential for applying the chain rule effectively. In composite functions, like \(y=\tanh(\cot x)\), the derivative of the hyperbolic tangent informs the rate of change of the outer function in relation to the inner function.