Problem 23 - Thomas Calculus in SI Units

Question

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\left(x^{2}+1\right) \operatorname{sech}(\ln x)$$ (Hint: Before differentiating, express in terms of exponentials and simplify.)

Step-by-Step Solution

Answer

The derivative \( \frac{dy}{dx} \) is 2.

Step 1: Recall Definition and Properties

Recall that the hyperbolic secant function is defined as \( \operatorname{sech}(u) = \frac{2}{e^u + e^{-u}} \). Also, remember the natural logarithm property: \( \ln(x^a) = a \ln(x) \). We'll use these to simplify the expression.

Step 2: Simplify the Hyperbolic Function

Given the function \( y = (x^2 + 1) \operatorname{sech}(\ln(x)) \), simplify using the definition: \( \operatorname{sech}(\ln(x)) = \frac{2}{e^{\ln(x)} + e^{-\ln(x)}} \). This simplifies to \( \operatorname{sech}(\ln(x)) = \frac{2}{x + \frac{1}{x}} = \frac{2x}{x^2 + 1} \). Substitute back into the expression: \( y = (x^2 + 1) \cdot \frac{2x}{x^2 + 1} \).

Step 3: Simplify the Expression Further

The equation becomes \( y = 2x \) since \( (x^2 + 1) \) and \( (x^2 + 1) \) cancel out in the numerator and denominator.

Step 4: Differentiate Simplified Function

Differentiate \( y = 2x \) with respect to \( x \). Using the power rule, the derivative of \( 2x \) is \( \frac{dy}{dx} = 2 \).

Step 5: Present the Final Derivative

The derived expression is simple: given \( y = 2x \, \frac{dy}{dx} = 2 \). This shows the derivative of the original function.

Key Concepts

Derivatives

When we talk about calculus, derivatives play a massive role in understanding how functions change. Derivatives measure the rate at which a function's value changes as its input changes. In this particular problem, we're asked to find how the function \( y \) changes concerning its variable, \( x \).
In simpler terms, a derivative will tell us how steeply the value of \( y \) rises or falls as \( x \) changes. It's like checking the speed of an object over time; the faster the change, the larger the derivative.
Derivatives have rules, often called calculus rules, to remember:

• Linearity: The derivative of \( a \cdot x + b \), where \( a \) and \( b \) are constants, is simply \( a \).
• Basic Function Rule: The derivative of \( x \) is 1, and so on for powers of \( x \).
• They help determine slopes of curves and optimize solutions for problems.

Thus, when differentiating a complex function, we often look for ways to simplify the components involved before employing these rules.

Hyperbolic functions

Hyperbolic functions are analogs of trigonometric functions but for hyperbolas rather than circles. They're often used in calculations involving hyperbolic geometry, and they appear in the problem of differentiating \( y = (x^2 + 1) \operatorname{sech}(\ln(x)) \).
**Understanding \( \operatorname{sech}(x) \):**
The hyperbolic secant, \( \operatorname{sech}(u) \), is an equivalent to the standard secant function but for hyperbolic sine and cosine. It's defined as: \( \operatorname{sech}(u) = \frac{2}{e^u + e^{-u}} \).
We use mathematic simplification and the properties of logarithms to rewrite the exponential terms, which greatly simplifies the derivative process.

• It takes advantage of algebraic identities and exponents' properties to make differentiation easier.
• In this problem, \( \operatorname{sech}(\ln(x)) \) becomes \( \frac{2x}{x^2 + 1} \), a simpler form to differentiate.

Conversely, remember that hyperbolic functions can often be expressed through exponentials, allowing easier differentiation.

Power rule

The power rule is a fundamental aspect of differentiation used frequently in calculus. It gives a straightforward way to find the derivative of powers of \( x \).
**Using the Power Rule:**
The general form of this rule is as follows: if \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \). It means we bring the exponent down as a coefficient and subtract one from the original exponent.

• For example, if \( f(x) = x^2 \) (as part of our original exercise), the derivative would be \( f'(x) = 2x^{2-1} = 2x \).

In the given solution, simplifying \( y = (x^2 + 1) \cdot \frac{2x}{x^2 + 1} \) to \( y = 2x \) allowed us to apply the power rule directly, ensuring a quick calculation of the derivative: \( \frac{dy}{dx} = 2 \).
This rule is extremely useful for polynomials and is often combined with other differentiation rules to solve more complex problems.