Problem 27
Question
Find the derivative. Simplify where possible. $$ f(x)=x \sinh x-\cosh x $$
Step-by-Step Solution
Verified Answer
The derivative is \(f'(x) = x \cosh x\).
1Step 1: Differentiate the First Term x sinh(x)
To differentiate the product of two functions, use the product rule. The product rule states that if you have two functions u(x) and v(x), then the derivative of their product is given by \((uv)' = u'v + uv'\). For \(x \sinh x\), let \(u = x\) and \(v = \sinh x\). Then, \(u' = 1\) and \(v' = \cosh x\). Apply the product rule to get: \(f_1'(x) = x \cosh x + 1 \cdot \sinh x\).
2Step 2: Differentiate the Second Term -cosh(x)
Differentiate \(-\cosh x\) with respect to \(x\). The derivative of \(\cosh x\) is \(\sinh x\), so \(\frac{d}{dx}(-\cosh x) = - \sinh x\). Thus, \(f_2'(x) = - \sinh x\).
3Step 3: Combine the Derivatives
Now, combine the results from Step 1 and Step 2. The derivative of the function \(f(x) = x \sinh x - \cosh x\) is the sum of its parts: \[f'(x) = (x \cosh x + \sinh x) + (-\sinh x)\].
4Step 4: Simplify the Expression
Combine like terms in the derivative. Notice that \(\sinh x\) and \(-\sinh x\) cancel each other out. Therefore, the simplified derivative is \(f'(x) = x \cosh x\).
Key Concepts
Product RuleHyperbolic FunctionsDifferentiationSimplification
Product Rule
In calculus, when you're asked to find the derivative of a product of two functions, the Product Rule is your go-to method. This handy rule helps us differentiate expressions where two functions are multiplied together.
If you have two functions, say \( u(x) \) and \( v(x) \), the derivative of their product \( u(x)v(x) \) is given by the rule:
Then, multiply each derivative with the respective other function and add the results.
In our exercise, for the term \( x \sinh x \), we set \( u = x \) and \( v = \sinh x \), hence \( u' = 1 \) and \( v' = \cosh x \).
Applying the Product Rule gives us \( x \cosh x + \sinh x \).
Understanding this rule is crucial for effectively tackling multi-part derivative problems.
If you have two functions, say \( u(x) \) and \( v(x) \), the derivative of their product \( u(x)v(x) \) is given by the rule:
- \[(uv)' = u'v + uv'\]
Then, multiply each derivative with the respective other function and add the results.
In our exercise, for the term \( x \sinh x \), we set \( u = x \) and \( v = \sinh x \), hence \( u' = 1 \) and \( v' = \cosh x \).
Applying the Product Rule gives us \( x \cosh x + \sinh x \).
Understanding this rule is crucial for effectively tackling multi-part derivative problems.
Hyperbolic Functions
Hyperbolic functions are similar to the trigonometric functions, but with some differences due to their relation to the geometry of hyperbolas.
The two most common hyperbolic functions are:
For differentiation, it is important to remember their derivatives: the derivative of \( \sinh x \) is \( \cosh x \), and the derivative of \( \cosh x \) is \( \sinh x \).
In the given problem, these properties are pivotal in differentiating \( x \sinh x \) and \( -\cosh x \). Understanding hyperbolic functions will enhance your calculus toolkit significantly.
The two most common hyperbolic functions are:
- Sinh (hyperbolic sine) \( \sinh x = \frac{e^x - e^{-x}}{2} \)
- Cosh (hyperbolic cosine) \( \cosh x = \frac{e^x + e^{-x}}{2} \)
For differentiation, it is important to remember their derivatives: the derivative of \( \sinh x \) is \( \cosh x \), and the derivative of \( \cosh x \) is \( \sinh x \).
In the given problem, these properties are pivotal in differentiating \( x \sinh x \) and \( -\cosh x \). Understanding hyperbolic functions will enhance your calculus toolkit significantly.
Differentiation
Differentiation is the process of finding the derivative, or the rate of change, of a function.
This is a fundamental concept in calculus and is used for finding slopes of curves, optimizing functions, and more.
In our exercise, differentiation involves differentiating each part of the function \( f(x) = x \sinh x - \cosh x \).
We used the Product Rule to handle the first part \( x \sinh x \) and direct differentiation for \( -\cosh x \).
Differentiating correctly ensures we understand how the function behaves at any given point.
This process helps us see patterns, make predictions, and solve complex real-world problems.
This is a fundamental concept in calculus and is used for finding slopes of curves, optimizing functions, and more.
In our exercise, differentiation involves differentiating each part of the function \( f(x) = x \sinh x - \cosh x \).
We used the Product Rule to handle the first part \( x \sinh x \) and direct differentiation for \( -\cosh x \).
Differentiating correctly ensures we understand how the function behaves at any given point.
This process helps us see patterns, make predictions, and solve complex real-world problems.
Simplification
Mathematical simplification is about making an expression more concise and efficient.
After finding the derivative in the given problem, the next step was to simplify. For instance, in the derivative:
Thus, the expression simplifies to \( x \cosh x \).
Simplification can make calculations easier and results clearer.By practicing simplification, you not only simplify the result, but increase your ability to spot these opportunities in future problems, reducing complexity.
After finding the derivative in the given problem, the next step was to simplify. For instance, in the derivative:
- \[(x \cosh x + \sinh x) - \sinh x\]
Thus, the expression simplifies to \( x \cosh x \).
Simplification can make calculations easier and results clearer.By practicing simplification, you not only simplify the result, but increase your ability to spot these opportunities in future problems, reducing complexity.
Other exercises in this chapter
Problem 26
Find the limit. $$\lim _{x \rightarrow \infty} \frac{2+10^{x}}{3-10^{x}}$$
View solution Problem 27
Differentiate the function. $$ F(t)=e^{t \sin 2 t} $$
View solution Problem 27
\(1-38=\) 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,
View solution Problem 27
Find the derivative of the function. Simplify where possible. \(y=x \sin ^{-1} x+\sqrt{1-x^{2}}\)
View solution