Chapter 6: Problem 14
In Problems 13-36, find Dy. $$ y=\cosh ^{2} x $$
Short Answer
Expert verified
\(dy = \sinh(2x)/2 \cdot dx\)
Step by step solution
01
Differentiate the Function with Respect to x
First, we want to find the derivative of the given function with respect to \( x \). We have \( y = \cosh^2 x \), which is a composite function since it involves \( \cosh x \) squared. We apply the chain rule, which requires us to take the derivative of the outer function and multiply it by the derivative of the inner function. Let \( u = \cosh x \). Then, \( y = u^2 \) with \( \frac{du}{dx} = \sinh x \). The derivative of \( y = u^2 \) is \( \frac{dy}{du} = 2u \). Substituting back, \( \frac{dy}{dx} = 2\cosh x \cdot \sinh x \).
02
Simplify the Expression
Using the identity \( \sinh(2x) = 2 \sinh x \cosh x \), we simplify \( \frac{dy}{dx} = 2\cosh x \sinh x \) to \( \sinh(2x) = 4 \sinh x \cosh x \). Therefore, \( \frac{dy}{dx} = \sinh(2x)/2 \).
03
Express Differentia Dy
The differential \( dy \) can be expressed using \( dy = \frac{dy}{dx} \cdot dx \). Substituting our derivative, \( dy = \sinh(2x)/2 \cdot dx \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differentiation Techniques
In calculus, differentiation is a way to determine how a function changes at any point, which is signified through a derivative. This process sheds light on the rate of change or the slope of the function at a specific point.
Differentiation techniques involve a variety of methods that can be used to find derivatives, each tailored to specific types of functions. Some of the fundamental differentiation techniques include:
Differentiation techniques involve a variety of methods that can be used to find derivatives, each tailored to specific types of functions. Some of the fundamental differentiation techniques include:
- Power Rule: Used for functions of the form \( f(x) = x^n \). The derivative is \( f'(x) = nx^{n-1} \).
- Product Rule: Used when you have a product of two functions. If \( u(x) \) and \( v(x) \) are functions, then \( (uv)' = u'v + uv' \).
- Quotient Rule: Used for a division of two functions. If \( u(x) \) and \( v(x) \) are functions, the derivative is \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).
Chain Rule
The chain rule is an essential differentiation technique used when dealing with composite functions. A composite function is a function within another function, and the chain rule helps us differentiate these by separating them into their components.When we have a function \( y = f(g(x)) \), the chain rule states that the derivative is given by \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). This means:
The chain rule is powerful and widely used because many real-world problems involve functions of other functions.
- First, differentiate the outer function (\( f \)), treating the inner function (\( g \)) as just a variable.
- Then, multiply this by the derivative of the inner function (\( g(x) \)).
The chain rule is powerful and widely used because many real-world problems involve functions of other functions.
Hyperbolic Functions
Hyperbolic functions are counterparts of the trigonometric functions, but they are defined using the exponential function \( e^x \). The most commonly used hyperbolic functions are \( \sinh x \) and \( \cosh x \), analogous to the sine and cosine in trigonometry.
These functions are defined as:
When differentiating hyperbolic functions, the derivatives are:
These functions are defined as:
- \( \sinh x = \frac{e^x - e^{-x}}{2} \)
- \( \cosh x = \frac{e^x + e^{-x}}{2} \)
When differentiating hyperbolic functions, the derivatives are:
- The derivative of \( \sinh x \) is \( \cosh x \).
- The derivative of \( \cosh x \) is \( \sinh x \).