Chapter 6: Problem 19
Derive the following derivative formulas given that \(d / d x(\cosh x)=\sinh x\) and \(d / d x(\sinh x)=\cosh x.\) $$d / d x(\operatorname{coth} x)=-\operatorname{csch}^{2} x$$
Short Answer
Expert verified
Answer: The derivative formula for coth(x) is: d(coth(x))/dx = -csch^2(x)
Step by step solution
01
Express coth(x) in terms of other hyperbolic functions
The function \(\operatorname{coth}(x)\) can be expressed as the quotient of \(\cosh(x)\) and \(\sinh(x)\):
$$
\operatorname{coth}(x) = \frac{\cosh(x)}{\sinh(x)}
$$
02
Apply the quotient rule
Now, we can use the quotient rule to find the derivative of \(\operatorname{coth}(x)\). The quotient rule states that if we have a function \(f(x)=\frac{u(x)}{v(x)}\), then the derivative is given by:
$$
f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^{2}}
$$
In our case, \(u(x)=\cosh(x)\) and \(v(x)=\sinh(x)\). We are given that \(u'(x)=\sinh(x)\) and \(v'(x)=\cosh(x)\). Plugging these values into the quotient rule, we get:
$$
\frac{d}{dx}(\operatorname{coth}(x)) = \frac{\sinh(x)\sinh(x) - \cosh(x)\cosh(x)}{[\sinh(x)]^{2}}
$$
03
Simplify the expression
To simplify the expression, we can combine terms in the numerator using the identity \(\cosh^2(x) - \sinh^2(x) = 1\). This leads to:
$$
\frac{d}{dx}(\operatorname{coth}(x)) = \frac{-1}{\sinh^2(x)}
$$
Now, we can rewrite the expression using the definition of the hyperbolic cosecant (csch) function as \(\operatorname{csch}(x) = \frac{1}{\sinh(x)}\):
$$
\frac{d}{dx}(\operatorname{coth}(x)) = -\operatorname{csch}^2(x)
$$
So, we have derived the given derivative formula for \(\operatorname{coth}(x)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Hyperbolic Functions
Hyperbolic functions are analogs of the trigonometric functions but for a hyperbola rather than a circle. They have similar names and properties, such as hyperbolic sine \( \sinh(x) \) and hyperbolic cosine \( \cosh(x) \):
The hyperbolic tangent \( \tanh(x) \) and cotangent \( \coth(x) \) are defined similarly to their trigonometric counterparts:
- Hyperbolic sine (sinh): \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
- Hyperbolic cosine (cosh): \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)
The hyperbolic tangent \( \tanh(x) \) and cotangent \( \coth(x) \) are defined similarly to their trigonometric counterparts:
- Hyperbolic tangent (tanh): \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \)
- Hyperbolic cotangent (coth): \( \coth(x) = \frac{\cosh(x)}{\sinh(x)} \)
- Hyperbolic cosecant (csch): \( \csch(x) = \frac{1}{\sinh(x)} \)
- \( \frac{d}{dx}(\cosh(x)) = \sinh(x) \)
- \( \frac{d}{dx}(\sinh(x)) = \cosh(x) \)
Derivative
The derivative is a tool in calculus that measures how a function changes as its input changes. It is often described as the "rate of change" or "slope" of the function. For example, the derivative of a function \( f(x) \) is denoted as \( f'(x) \) or \( \frac{df}{dx} \).
For hyperbolic functions, knowing the derivatives is crucial for differentiating more complex expressions. For example:
Mastering derivatives is fundamental for understanding more sophisticated applications of calculus, like solving equations or modeling real-world scenarios.
For hyperbolic functions, knowing the derivatives is crucial for differentiating more complex expressions. For example:
- The derivative of the hyperbolic sine \( \sinh(x) \) is \( \cosh(x) \).
- Conversely, the derivative of the hyperbolic cosine \( \cosh(x) \) is \( \sinh(x) \).
Mastering derivatives is fundamental for understanding more sophisticated applications of calculus, like solving equations or modeling real-world scenarios.
Quotient Rule
The quotient rule is a method for finding the derivative of a quotient of two differentiable functions. Suppose we have a function \( f(x) = \frac{u(x)}{v(x)} \), where both \( u(x) \) and \( v(x) \) are differentiable.
The quotient rule states:\[\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \\]This tells us how to calculate the derivative by using the derivatives of the numerator \( u(x) \) and the denominator \( v(x) \).
Here's a breakdown of the steps:
The quotient rule states:\[\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \\]This tells us how to calculate the derivative by using the derivatives of the numerator \( u(x) \) and the denominator \( v(x) \).
Here's a breakdown of the steps:
- Step 1: Differentiate the numerator to get \( u'(x) \).
- Step 2: Differentiate the denominator to get \( v'(x) \).
- Step 3: Apply the formula by plugging in \( u'(x) \), \( v(x) \), and \( v'(x) \).