Problem 25 - Calculus of a Single Variable

Question

In Exercises 23–32, find the derivative of the function. $$ y=\operatorname{sech}\left(5 x^{2}\right) $$

Step-by-Step Solution

Answer

The derivative of the function \(y=\operatorname{sech}\left(5 x^{2}\right)\) is \(dy/dx = -10x \operatorname{sech}(5x^2)\operatorname{tanh}(5x^2)\).

Step 1: Identify the function

Identify the function given, which is \(y=\operatorname{sech}\left(5 x^{2}\right)\). Here, \(u = 5x^2\) is the function inside the hyperbolic secant function.

Step 2: Differentiate the inner function

Apply the power rule to differentiate the inner function \(u\). The derivative of \(u\), denoted \(u'\), is \(10x\).

Step 3: Differentiate the outer function

Differentiate the outer function, the hyperbolic secant of \(u\). Note that the derivative of \(\operatorname{sech}(x)\) is \(-\operatorname{sech}(x)\operatorname{tanh}(x)\). After replacing \(x\) with \(u\), the derivative is \(-\operatorname{sech}(u)\operatorname{tanh}(u)\).

Step 4: Apply the chain rule

By the chain rule, if \(y\) is a composition of two functions \(u\) and \(v\) such that \(y = v(u(x))\), then the derivative of \(y\) with respect to \(x\) is given by \(dy/dx = v'(u) * u'\). Substituting \(v'(u) = -\operatorname{sech}(u)\operatorname{tanh}(u)\) and \(u' = 10x\), gives \(dy/dx = -10x \operatorname{sech}(u)\operatorname{tanh}(u)\).

Step 5: Substitute u back into the equation

Finally, you substitute \(u=5x^2\) back into the equation. So, \(dy/dx = -10x \operatorname{sech}(5x^2)\operatorname{tanh}(5x^2)\).

Key Concepts

Chain Rule

When dealing with composite functions, the chain rule is an essential tool in calculus for finding the derivative. It allows us to differentiate a function that is composed of two or more functions nestled within each other. For example, if you have a function expressed as \(y = v(u(x))\), the chain rule states that \(\frac{dy}{dx} = v'(u) \cdot u'\).
Here:

• \(u(x)\) is the inner function, which needs to be differentiated first.
• \(v(u)\) is the outer function, which is differentiated next, evaluated at \(u(x)\).

In the problem given, the inner function is \(u = 5x^2\), and the outer function is \(\operatorname{sech}(u)\). By applying the chain rule, you first differentiate \(u\) to get \(u' = 10x\). Then, you differentiate the outer function with respect to \(u\). Finally, multiply the derivatives of the inner and outer functions together to get the full derivative.

Hyperbolic Functions

Hyperbolic functions, akin to trigonometric functions, often appear in various calculus problems. They are best understood as similar, yet distinct, from their trigonometric counterparts.
The hyperbolic secant function is given by \(\operatorname{sech}(x) = \frac{2}{e^x + e^{-x}}\). Its derivative, however, has a unique property: \(\frac{d}{dx} [\operatorname{sech}(x)] = -\operatorname{sech}(x) \cdot \operatorname{tanh}(x)\).

• The \(\operatorname{tanh}(x)\), or hyperbolic tangent, is defined as \(\operatorname{tanh}(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}\).

Understanding hyperbolic functions' derivatives is critical, because they incorporate combinations of \(e^x\) and \(e^{-x}\) expressions. This exercise includes \(\operatorname{sech}\), where our task was to find its derivative, which involves \(\operatorname{tanh}(x)\), exploitable through the hyperbolic identity.

Power Rule

The power rule is the go-to differentiation rule for functions of the form \(x^n\), where \(n\) is any real number. It allows us to efficiently find derivatives by simplifying the differentiation process. According to the rule: for \(f(x) = x^n\), the derivative \(f'(x) = n \cdot x^{n-1}\).
In our example, the inner function \(u(x) = 5x^2\) follows the format of the power rule. Here, \(n = 2\). By applying the power rule, we differentiate \(5x^2\) to obtain \(u'(x) = 10x\).

• Start by bringing down the exponent (in our case, 2).
• Multiply it by the coefficient (here, 5).
• Subtract 1 from the exponent to simplify \(x^1\).

This illustrates the power rule's usefulness in rapidly handling polynomial expressions and setting up more complex derivatives.