Question

Derive the following derivative formulas given that \(\frac{d}{d x}(\cosh x)=\sinh x\)and\(\frac{d}{d x}(\sinh x)=\cosh x\). $$\frac{d}{d x}(\operatorname{sech} x)=-\operatorname{sech} x \tanh x$$

Short answer

Answer: The derivative formula for the sech function is: \(\frac{d}{dx}(\operatorname{sech}(x)) = -\operatorname{sech}(x) \tanh(x)\).

Step-by-step solution

Define the sech function in terms of cosh

To express \(\operatorname{sech}(x)\) in terms of \(\cosh(x)\), we should recall that \(\operatorname{sech}(x)\) is defined as the reciprocal of \(\cosh(x)\). Thus, it can be represented as: $$\operatorname{sech}(x)=\frac{1}{\cosh(x)}$$

Differentiate the sech function

Now, we need to differentiate the sech function with respect to x. To accomplish this, we will use the quotient rule: $$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$$ In our case, \(u=1\) and \(v=\cosh(x)\). Therefore, $$\frac{d}{d x}\left(\operatorname{sech} x\right) = \frac{\cosh(x)\frac{d}{dx}(1) - 1 \frac{d}{dx}(\cosh x)}{\cosh^2(x)}$$

Use the given derivative formulas

Next, we will use the given derivative formulas \(\frac{d}{dx}(\cosh(x)) = \sinh(x)\) and \(\frac{d}{dx}(1) = 0\) to simplify the expression: $$\frac{d}{d x}(\operatorname{sech} x) = \frac{\cosh(x)(0) - 1(\sinh x)}{\cosh^2(x)} = \frac{-\sinh(x)}{\cosh^2(x)}$$

Express the result in terms of sech and tanh functions

We can now rewrite the result in terms of \(\operatorname{sech}(x)\) and \(\tanh(x)\). Recall that \(\tanh(x) = \frac{\sinh(x)}{\cosh(x)}\), and \(\operatorname{sech}(x) = \frac{1}{\cosh(x)}\). Substituting these definitions, we get: $$\frac{d}{d x}(\operatorname{sech} x) = -\operatorname{sech}(x)\cosh(x) \cdot \tanh(x) = -\operatorname{sech}(x) \tanh(x)$$ Thus, the derivative formula for the sech function is: $$\frac{d}{d x}(\operatorname{sech} x)=-\operatorname{sech} x \tanh x$$

Key concepts

Understanding the Sech Function

The sech function, written as \(\operatorname{sech}(x)\), is one of the hyperbolic functions and is defined as the reciprocal of the hyperbolic cosine, \(\cosh(x)\). In other words:

• \(\operatorname{sech}(x) = \frac{1}{\cosh(x)}\)

Understanding this definition is crucial because it allows us to explore its properties and relationships with other hyperbolic functions. The sech function is specifically useful in various fields, including calculus and differential equations, due to its unique behavior and symmetry. For instance, understanding the sech function helps in determining the shape of curves and surfaces in mathematical models.

Applying the Quotient Rule

To find the derivative of \(\operatorname{sech}(x)\), using the quotient rule is essential. The quotient rule is a method for differentiating functions that are expressed as a quotient (or division) of two functions. The rule is expressed as:

• \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\)

In our case, \(u = 1\) and \(v = \cosh(x)\). Applying the quotient rule means differentiating the numerator and denominator separately. Since \(\frac{d}{dx}(1) = 0\) and \(\frac{d}{dx}(\cosh(x)) = \sinh(x)\), the derivative using the quotient rule becomes: \[ \frac{\cosh(x) \cdot 0 - 1 \cdot \sinh(x)}{\cosh^2(x)} = \frac{-\sinh(x)}{\cosh^2(x)} \] By applying the quotient rule effectively, we've rewritten the derivative of the reciprocal hyperbolic cosine function, giving us a clear expression to work with.

Using Cosh and Sinh Derivatives

Knowing the derivatives of hyperbolic sine and cosine functions is crucial when calculating the derivative of \(\operatorname{sech}(x)\). Here are the given derivative formulas:

• \(\frac{d}{dx}(\cosh(x)) = \sinh(x)\)
• \(\frac{d}{dx}(\sinh(x)) = \cosh(x)\)

These fundamental rules simplify the differentiation process. For example, when we applied the quotient rule to \(\operatorname{sech}(x)\) and reached \(-\frac{\sinh(x)}{\cosh^2(x)}\), we utilized these derivatives. Transforming this expression using the identity \(\tanh(x) = \frac{\sinh(x)}{\cosh(x)}\), we substitute the definitions:

• \(\operatorname{sech}(x) = \frac{1}{\cosh(x)}\)
• \(\tanh(x) = \frac{\sinh(x)}{\cosh(x)}\)

resulting in the compact expression for the derivative: \[ \frac{d}{d x}(\operatorname{sech} x) = -\operatorname{sech}(x) \tanh(x) \] Understanding and using these hyperbolic derivatives helps make complex calculus problems manageable and comprehensible.