Problem 27 - Calculus Early Transcendentals

Question

Compute \(dy/dx\) for the following functions. $$y=\ln \operatorname{sech} 2 x$$

Step-by-Step Solution

Answer

Answer: The derivative of the function $$y = \ln(\operatorname{sech}(2x))$$ with respect to x is $$\frac{dy}{dx} = -2\operatorname{tanh}(2x)$$.

Step 1: Recall the chain rule and the derivative of the natural logarithm

To find the derivative of the given function, we'll need the chain rule, which is: $$\frac{d}{dx}\big[f(g(x))\big] = f'(g(x)) \cdot g'(x)$$ And the derivative of the natural logarithm function is: $$\frac{d}{dx}\big[\ln(x)\big] = \frac{1}{x}$$

Step 2: Differentiate the function using the chain rule

Applying the chain rule to $$y = \ln \operatorname{sech} 2 x$$, we differentiate the outer function (natural logarithm) first, and then the inner function (secant hyperbolic): $$\frac{dy}{dx} = \frac{d}{dx}\big[\ln(\operatorname{sech}(2x))\big] = \frac{1}{\operatorname{sech}[2x]} \cdot \frac{d}{dx}\big(\operatorname{sech}(2x)\big)$$

Step 3: Recall the derivative of the secant hyperbolic function, and any needed identities

The derivative of the secant hyperbolic function is given by: $$\frac{d}{dx}\big[\operatorname{sech}(x)\big] = -\operatorname{sech}(x) \operatorname{tanh}(x)$$ Also recall that hyperbolic functions are related to each other by the identity: $$\operatorname{sech}^{2}(x) + \operatorname{tanh}^{2}(x) = 1$$

Step 4: Differentiate the secant hyperbolic function and substitute for the tangent hyperbolic function

Differentiating the secant hyperbolic function, and substituting for the tangent hyperbolic function, we get: $$\frac{d}{dx}\big[\operatorname{sech}(2x)\big] = -2\operatorname{sech}(2x) \operatorname{tanh}(2x)$$

Step 5: Substitute the derivative of the secant hyperbolic function into the chain rule expression

Substitute the derivative of the secant hyperbolic function into the expression we obtained in Step 2: $$\frac{dy}{dx} = \frac{1}{\operatorname{sech}[2x]} \cdot (-2\operatorname{sech}(2x) \operatorname{tanh}(2x))$$

Step 6: Simplify the expression for the derivative

Finally, simplifying the expression for the derivative we obtained in Step 5, we get: $$\frac{dy}{dx} = -2\operatorname{tanh}(2x)$$ So the derivative of $$y = \ln \operatorname{sech} 2 x$$ with respect to x is: $$\frac{dy}{dx} = -2\operatorname{tanh}(2x)$$

Key Concepts

chain rule

The chain rule is a powerful tool in calculus used to find the derivative of a composition of functions. It states that if you have two functions, an outer function \(f\) and an inner function \(g\), the derivative of their composition \(f(g(x))\) is given by

• Taking the derivative of the outer function while keeping the inner function unchanged, and
• Multiplying this by the derivative of the inner function.

This can be written mathematically as: \[\frac{d}{dx}\big[f(g(x))\big] = f'(g(x)) \cdot g'(x)\]In the exercise, the outer function is the natural logarithm \(\ln(x)\), and the inner function is the hyperbolic secant \(\operatorname{sech}(2x)\). First, we find the derivative of \(\ln(x)\), which is \(1/x\), and then multiply it by the derivative of \(\operatorname{sech}(2x)\). It's a step-by-step method that ensures we differentiate correctly each part of the composed function.

natural logarithm

The natural logarithm is a fundamental mathematical function often denoted as \(\ln(x)\). It is the inverse of the exponential function with base \(e\) (Euler's number, approximately 2.718). It transforms multiplicative processes into additive ones, making many types of calculations simpler.

• The derivative of the natural logarithm, \(\ln(x)\), with respect to \(x\) is \(1/x\).
• This property is extremely useful in differentiating functions where the logarithm is involved.

In the given exercise, we differentiate \(\ln(\operatorname{sech}(2x))\) by applying the derivative \(1/\operatorname{sech}(2x)\), which is then multiplied by the derivative of the inner function according to the chain rule. Understanding the natural logarithm and its properties, such as its differentiability, is essential for solving problems involving exponential growth, decay, and other natural phenomena.

hyperbolic functions

Hyperbolic functions, similar in concept to trigonometric functions, are functions that relate to hyperbolas much in the way that trigonometric functions relate to circles. One of these functions is the hyperbolic secant, \(\operatorname{sech}(x)\), defined as:\[\operatorname{sech}(x) = \frac{2}{e^x + e^{-x}}\]The derivative of \(\operatorname{sech}(x)\) is given by:\[\frac{d}{dx}\big[\operatorname{sech}(x)\big] = -\operatorname{sech}(x) \operatorname{tanh}(x)\]

• \(\operatorname{tanh}(x)\) is the hyperbolic tangent function, defined as \(\operatorname{tanh}(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}\).
• The identity \(\operatorname{sech}^2(x) + \operatorname{tanh}^2(x) = 1\) is a useful relationship between hyperbolic functions.

In the exercise, we use these properties to derive \(\frac{d}{dx}\big[\operatorname{sech}(2x)\big]\), making use of the chain rule and simplifying the derivative expression. Mastery of hyperbolic functions often aids in solving complex calculus problems related to real-world phenomena, like catenary shapes in architecture or the behavior of electrical circuits.