Problem 44 - Calculus Single Variable

Question

Differentiate the given expression with respect to \(x\). $$ x \operatorname{sech}(x) $$

Step-by-Step Solution

Answer

The derivative is \(\operatorname{sech}(x) - x \operatorname{sech}(x)\operatorname{tanh}(x)\).

Step 1: Identify the Differentiation Rule

We need to differentiate the expression \(x \operatorname{sech}(x)\). This requires the product rule for differentiation since it is a product of two functions: \(u(x) = x\) and \(v(x) = \operatorname{sech}(x)\). The product rule states that \((uv)' = u'v + uv'\).

Step 2: Differentiate Each Function

First, differentiate \(u(x) = x\) with respect to \(x\). The derivative \(u'(x) = 1\).Next, differentiate \(v(x) = \operatorname{sech}(x)\) with respect to \(x\). Remember the derivative of \(\operatorname{sech}(x)\) is \(-\operatorname{sech}(x)\operatorname{tanh}(x)\). So, \(v'(x) = -\operatorname{sech}(x)\operatorname{tanh}(x)\).

Step 3: Apply the Product Rule

Using the product rule \((uv)' = u'v + uv'\), substitute the derivatives we found:1. \( u'(x) = 1\) and \(v(x) = \operatorname{sech}(x)\)2. \( u(x) = x\) and \(v'(x) = -\operatorname{sech}(x)\operatorname{tanh}(x)\)Therefore, the derivative is:\[(x \operatorname{sech}(x))' = 1 \cdot \operatorname{sech}(x) + x \cdot (-\operatorname{sech}(x)\operatorname{tanh}(x))\]

Step 4: Simplify the Expression

Simplify the expression from applying the product rule:\[(x\operatorname{sech}(x))' = \operatorname{sech}(x) - x \operatorname{sech}(x)\operatorname{tanh}(x)\]This combines into a single expression for the derivative. No further simplification is needed.

Key Concepts

Calculus

Calculus is a fundamental branch of mathematics that deals with rates of change and the accumulation of quantities. In essence, it focuses on differentiation and integration. Differentiation is the process of finding the derivative, which represents the rate at which a function changes at any given point. The derivative is a core concept in calculus and serves as a foundational tool in fields like physics, engineering, and economics.

• Rate of Change: Differentiation helps in determining how one quantity changes with respect to another, signifying the concept of a rate of change.
• Function Behavior: Through derivatives, we can understand the behavior of functions, such as finding local maxima and minima, and analyzing graph shapes.

In our exercise, we were tasked with differentiating a product of functions using calculus rules. The solution required the application of the product rule, a specialized technique in differentiation, to unravel the rate at which our expression changes.

Product Rule

The product rule is a key formula in differentiation used to find the derivative of the product of two differentiable functions. In simple terms, when you have two functions multiplied together, their derivative is not just the derivative of one times the other but requires a specific approach. The product rule formula states that for two functions, \(u(x)\) and \(v(x)\), the derivative is:

• \((uv)' = u'v + uv'\)

Here’s a breakdown of the product rule application:

• First, differentiate the first function \(u(x)\) to get \(u'(x)\).
• Next, multiply \(u'(x)\) with the second function \(v(x)\).
• Then, differentiate the second function \(v(x)\) to get \(v'(x)\).
• Multiply \(v'(x)\) with the original first function \(u(x)\).
• Add the results from the two products.

In our task, the functions were \( u(x) = x \) and \( v(x) = \operatorname{sech}(x) \). Applying the product rule, we calculated the derivative of \(x \operatorname{sech}(x)\) to find the desired rate of change of this expression.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but for a hyperbola, as opposed to a circle. They have applications in various aspects of mathematics, including calculus, and physics. The six basic hyperbolic functions are similar to their trigonometric counterparts: sinh, cosh, tanh, csch, sech, and coth. In our exercise, \(\operatorname{sech}(x)\) appears, which is related to the hyperbolic cosine function.

• Understanding \(\operatorname{sech}(x)\):
  • \(\operatorname{sech}(x)\) stands for hyperbolic secant, defined as \(1/\cosh(x)\), where \(\cosh(x)\) is \((e^x + e^{-x})/2\).
  • The derivative of \(\operatorname{sech}(x)\) is \(-\operatorname{sech}(x)\operatorname{tanh}(x)\).
• Useful Properties:
  • Hyperbolic functions are instrumental in solving certain types of differential equations.
  • They often appear in the description of the shape of a hanging cable, known as a catenary.

Mastering hyperbolic functions can enhance your ability to tackle complex integrals and differential equations, just as we employed \(\operatorname{sech}(x)\) and its derivative in our differentiation exercise.