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Magazine Chapter 6: Problem 23
Compute \(d y / d x\) for the following functions. \(y=\cosh ^{2} x\)
Short Answer
Expert verified
Answer: The derivative of the function \(y = \cosh^2 x\) is \(dy/dx = 2\cosh x \cdot \sinh x\).
Step by step solution
01
Identify the inner and outer functions
The given function \(y = \cosh^2 x\) can be written as a composition of two functions: an outer function \(u^2\) and an inner function \(u = \cosh x\).
02
Apply the Chain Rule
According to the chain rule, if \(y = f(g(x))\), then \(y' = f'(g(x)) \cdot g'(x)\). In our case, \(f(u) = u^2\) and \(g(x) = \cosh x\). Therefore, we need to find the derivatives of both these functions: \(f'(u)\) and \(g'(x)\).
03
Compute the derivative of the outer function
To find the derivative of the outer function \(f(u) = u^2\), we can simply apply the power rule for differentiation: \((u^n)' = nu^{n - 1}\). In our case, \(n = 2\), so we get:
\(f'(u) = 2u^{2 - 1} = 2u\).
04
Compute the derivative of the inner function
To find the derivative of the inner function \(g(x) = \cosh x\), we can use the fact that \(\cosh x\) is equal to \((e^x + e^{-x})/2\). Applying the chain rule and the fact that \((e^x)' = e^x\) and \((e^{-x})' = -e^{-x}\), we find the derivative of \(\cosh x\) as follows:
\(g'(x) = \sinh x\)
05
Use the chain rule formula to find the derivative of the composed function
Now we can substitute the derivatives found above into the chain rule formula:
\(y' = f'(g(x)) \cdot g'(x) \Rightarrow y' = (2u) \cdot \sinh x\)
Since the inner function \(u = \cosh x\), we can substitute it back:
\(y' = 2(\cosh x) \cdot \sinh x\)
Therefore, the derivative of \(y = \cosh^2 x\) is:
\(dy/dx = 2\cosh x \cdot \sinh x\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cosh function
The hyperbolic cosine function, denoted as \( \cosh x \), is a fundamental hyperbolic function. It's defined as the average of exponential growth and decay: \( \cosh x = \frac{e^x + e^{-x}}{2} \). Similar to the cosine function of circular trigonometry, \( \cosh x \) has properties that make it significant in calculus and differential equations.
Understanding this function involves knowing its relationship with exponential functions:
Understanding this function involves knowing its relationship with exponential functions:
- The function is even, meaning \( \cosh(-x) = \cosh x \).
- As \( x \to \infty \), \( \cosh x \to \infty \), and as \( x \to -\infty \), \( \cosh x \to \infty \) due to its symmetric nature.
Chain rule
The chain rule is a crucial tool in differentiation, especially when dealing with composite functions. A composite function is a function of a function, like \( y = \cosh^2 x \), which involves both the outer and inner functions.
To apply the chain rule, follow these steps:
To apply the chain rule, follow these steps:
- Identify the outer function and the inner function. Here, the outer function is \( u^2 \) and the inner is \( u = \cosh x \).
- Differentiate the outer function with respect to the inner function. Using the power rule \( (u^n)' = nu^{n-1} \), we find \( f'(u) = 2u \).
- Differentiate the inner function, which is \( g(x) = \cosh x \). We get \( g'(x) = \sinh x \), using properties of hyperbolic functions.
Hyperbolic functions
Hyperbolic functions, including \( \cosh x \) and \( \sinh x \), are analogs to the trigonometric functions but for a hyperbola. They come naturally out of the solutions of hyperbolic geometry and the Lorentz transformations in relativity. Here are some key points about them:
- \( \sinh x = \frac{e^x - e^{-x}}{2} \)
- Unlike sine and cosine, \( \cosh x \) and \( \sinh x \) do not repeat or cycle.
- Derivatives: \( \frac{d}{dx}\cosh x = \sinh x \) and \( \frac{d}{dx}\sinh x = \cosh x \)
Derivative calculation
Calculating derivatives is a fundamental process in calculus. It involves understanding how a function changes as its input changes, which leads to finding slopes at any given point of a function. For the function \( y = \cosh^2 x \), performing differentiation involves using concepts we've previously discussed:
- Use the chain rule for \( y = f(g(x)) \) where \( f'(g(x)) \) and \( g'(x) \) are known.
- In our case: \( f(u) = u^2 \) leading to \( f'(u) = 2u \), and \( g(x) = \cosh x \) giving \( g'(x) = \sinh x \).
- Combine the derivatives using the chain rule: \( y' = 2(\cosh x)(\sinh x) \), resulting in the overall derivative \( dy/dx = 2\cosh x \cdot \sinh x \).
