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Magazine Chapter 8: Problem 12
Exer. \(3-26:\) Find \(f^{\prime}(x)\) if \(f(x)\) is the given expression. $$ \sqrt{\operatorname{sech} 5 x} $$
Short Answer
Expert verified
The derivative is \( f'(x) = -\frac{5\operatorname{sech}(5x)\operatorname{tanh}(5x)}{2\sqrt{\operatorname{sech}(5x)}} \).
Step by step solution
01
Recognize Function Composition
The function \( f(x) = \sqrt{\operatorname{sech} 5x} \) is a composition of functions. Identify the outer function as \( u(x) = \sqrt{x} \) and the inner function as \( v(x) = \operatorname{sech}(5x) \). This will require the use of the chain rule to differentiate.
02
Differentiate the Outer Function
Differentiate the outer function \( u(x) = \sqrt{x} \) with respect to \( x \), using the derivative \( \frac{d}{dx}\sqrt{x} = \frac{1}{2\sqrt{x}} \). This gives us: \( u'(v(x)) = \frac{1}{2\sqrt{\operatorname{sech} 5x}} \).
03
Simplify the Inner Function
Recall that the hyperbolic secant function is defined as \( \operatorname{sech}(x) = \frac{2}{e^x + e^{-x}} \). However, for differentiation, we will employ its derivative directly without explicit simplification into trigonometric or hyperbolic identities.
04
Differentiate the Inner Function
Differentiate the inner function \( v(x) = \operatorname{sech}(5x) \). Use the derivative \( \frac{d}{dx}\operatorname{sech}(x) = -\operatorname{sech}(x)\operatorname{tanh}(x) \), and account for the chain rule when differentiating \( 5x \), which results in \( 5\operatorname{sech}(5x)\operatorname{tanh}(5x) \).
05
Apply the Chain Rule
Combine the results from Steps 2 and 4 using the chain rule. The chain rule gives \( f'(x) = u'(v(x)) \cdot v'(x) \). Substitute the derivatives from prior steps: \[ f'(x) = \frac{1}{2\sqrt{\operatorname{sech} 5x}} \cdot (-5) \operatorname{sech}(5x)\operatorname{tanh}(5x). \]
06
Simplify the Expression
Simplify the expression: \[ f'(x) = -\frac{5\operatorname{sech}(5x)\operatorname{tanh}(5x)}{2\sqrt{\operatorname{sech}(5x)}}. \] This is the simplified form of the derivative.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hyperbolic Functions
Hyperbolic functions like the hyperbolic secant (\( \operatorname{sech}(x) \)) are analogous to trigonometric functions but relate to hyperbolas instead of circles. These functions are based on the exponential function \( e^x \) and have important applications in many areas of mathematics, including calculus and differential equations.
- The hyperbolic secant is defined as \( \operatorname{sech}(x) = \frac{2}{e^x + e^{-x}} \).
- Like trigonometric functions, hyperbolic functions have derivatives that are useful for solving calculus problems.
- The derivative of \( \operatorname{sech}(x) \) is different but shares some functional similarities with its trig counterpart.
Differentiation
Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes at any given point. In this exercise, differentiation plays a crucial role in determining the slope of the tangent to the curve defined by \( f(x) = \sqrt{\operatorname{sech}(5x)} \). Here are some essential aspects:
- The differentiation of a composite function, like in our example, requires careful application of the chain rule.
- The chain rule helps break down the process into manageable steps by differentiating the outer and inner functions successively.
- Correctly deriving hyperbolic functions often involves remembering specific derivative formulas, such as \( \frac{d}{dx}\operatorname{sech}(x) = -\operatorname{sech}(x)\operatorname{tanh}(x) \).
Function Composition
Function composition occurs when you have a function within another function, a situation frequently found in calculus problems. In \( f(x) = \sqrt{\operatorname{sech}(5x)} \), we see two layers of functions:
- The outer function is \( u(x) = \sqrt{x} \), which means we're applying a square root.
- The inner function is \( v(x) = \operatorname{sech}(5x) \).
