Problem 29
Question
Find the derivative. Simplify where possible. $$ h(x)=\ln (\cosh x) $$
Step-by-Step Solution
Verified Answer
The derivative of \( h(x) = \ln(\cosh x) \) is \( \tanh x \).
1Step 1: Differentiate the Logarithmic Function
To find the derivative of the function \( h(x) = \ln(\cosh x) \), we apply the chain rule. The derivative of \( \ln(u) \) with respect to \( x \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = \cosh x \).
2Step 2: Differentiate the Hyperbolic Function
Next, we differentiate \( u = \cosh x \). The derivative of \( \cosh x \) is \( \sinh x \). So, \( \frac{du}{dx} = \sinh x \).
3Step 3: Apply the Chain Rule
Using the chain rule in Step 1, we get \( h'(x) = \frac{1}{\cosh x} \cdot \sinh x \).
4Step 4: Simplify the Expression
The expression \( \frac{\sinh x}{\cosh x} \) simplifies to \( \tanh x \) based on the identity \( \tanh x = \frac{\sinh x}{\cosh x} \). Thus, the derivative \( h'(x) = \tanh x \).
Key Concepts
Chain RuleHyperbolic FunctionsSimplification
Chain Rule
The chain rule is a crucial concept in calculus that allows us to find the derivative of composite functions. When a function is nested inside another, like the function we have with a logarithm and a hyperbolic function, the chain rule becomes extremely useful.
To apply the chain rule, we first identify the inner function and the outer function. In our case, the outer function is the natural logarithm, \( \ln(u) \), and the inner function is \( \cosh x \).
\[\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx}.\] This concept helps us transition step by step from differentiating simpler functions to tackling more complex combinations of functions.
To apply the chain rule, we first identify the inner function and the outer function. In our case, the outer function is the natural logarithm, \( \ln(u) \), and the inner function is \( \cosh x \).
- First, differentiate the outer function: The derivative of \( \ln(u) \) is \( \frac{1}{u} \).
- Next, differentiate the inner function \( u = \cosh x \): The derivative is \( \sinh x \).
\[\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx}.\] This concept helps us transition step by step from differentiating simpler functions to tackling more complex combinations of functions.
Hyperbolic Functions
Hyperbolic functions, analogous to trigonometric functions, are very useful in many areas of mathematics and science. They are defined using the exponential function and form identities much like trigonometric identities.
One key hyperbolic function is the hyperbolic cosine, \( \cosh x \), which is defined as:
\[\cosh x = \frac{e^x + e^{-x}}{2}.\]
When differentiating hyperbolic functions, it's essential to know their derivatives by heart for quick calculations. For \( \cosh x \), its derivative is the hyperbolic sine function:
\[\frac{d}{dx} \cosh x = \sinh x.\]
Hyperbolic functions also possess properties and identities that are often symmetrical to their trigonometric counterparts. One might think of them as another fascinating layer of the broader study of calculus functions.
One key hyperbolic function is the hyperbolic cosine, \( \cosh x \), which is defined as:
\[\cosh x = \frac{e^x + e^{-x}}{2}.\]
When differentiating hyperbolic functions, it's essential to know their derivatives by heart for quick calculations. For \( \cosh x \), its derivative is the hyperbolic sine function:
\[\frac{d}{dx} \cosh x = \sinh x.\]
Hyperbolic functions also possess properties and identities that are often symmetrical to their trigonometric counterparts. One might think of them as another fascinating layer of the broader study of calculus functions.
Simplification
Simplification in calculus is the process of making an expression or equation more manageable and easier to understand. When we deal with derivatives, simplification can bring clarity and highlight key insights.
In our example, after applying the chain rule, we found the expression \( \frac{\sinh x}{\cosh x} \). This can be simplified using the identity for the hyperbolic tangent:
\[\tanh x = \frac{\sinh x}{\cosh x}.\]
This transformation from the fraction to \( \tanh x \) not only makes the expression neater but also immediately tells us more about the function's behavior. Such simplifications are fundamental—cutting down complexity and often revealing the relationships between various mathematical functions.
When working through derivatives, always remember that identifying these simplifications can make solutions more elegant and understandable.
In our example, after applying the chain rule, we found the expression \( \frac{\sinh x}{\cosh x} \). This can be simplified using the identity for the hyperbolic tangent:
\[\tanh x = \frac{\sinh x}{\cosh x}.\]
This transformation from the fraction to \( \tanh x \) not only makes the expression neater but also immediately tells us more about the function's behavior. Such simplifications are fundamental—cutting down complexity and often revealing the relationships between various mathematical functions.
When working through derivatives, always remember that identifying these simplifications can make solutions more elegant and understandable.
Other exercises in this chapter
Problem 28
Find the limit. $$\lim _{x \rightarrow 2^{-}} e^{3 /(2-x)}$$
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Differentiate the function. $$ y=\ln \left|2-x-5 x^{2}\right| $$
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\(1-38=\) Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply,
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Find the derivative of the function. Simplify where possible. \(y=\arccos \left(\frac{b+a \cos x}{a+b \cos x}\right), \quad 0 \leqslant x \leqslant \pi, a>b>0\)
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