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Find \(d y / d x\) $$y=\sinh (\cos 3 x)$$

Short Answer

Expert verified
\(\frac{dy}{dx} = -3 \sin 3x \cdot \cosh(\cos 3x)\)

Step by step solution

01

Identify the Composition of Functions

The function given is \(y = \sinh(\cos 3x)\). This is a composition of the hyperbolic sine function \(\sinh(u)\) where \(u = \cos 3x\). Recognize that we will need to use the chain rule to differentiate.
02

Differentiate the Outer Function

The derivative of \(\sinh(u)\) with respect to \(u\) is \(\cosh(u)\). Therefore, differentiate \(\sinh(\cos 3x)\) with respect to \(\cos 3x\), resulting in \(\cosh(\cos 3x)\).
03

Differentiate the Inner Function

The inner function is \(u = \cos 3x\). The derivative \(\frac{d}{dx}(\cos 3x)\) is \(-\sin 3x\). Using the chain rule, account for the inner function derivative when differentiating \(\sinh(\cos 3x)\).
04

Apply the Chain Rule

Using the chain rule \(\frac{d}{dx}[\sinh(\cos 3x)] = \cosh(\cos 3x) \cdot (-\sin 3x) \cdot \frac{d}{dx}(3x)\). The derivative of \(3x\) is \(3\). So, commbine results as \(-3 \sin 3x \cdot \cosh(\cos 3x)\).
05

Final Expression for \(\frac{dy}{dx}\)

The final derivative after using the chain rule and simplifying is \(\frac{dy}{dx} = -3 \sin 3x \cdot \cosh(\cos 3x)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation
Differentiation is a core concept in calculus that allows us to determine the rate at which a function is changing at any point. In simpler terms, it gives us the slope of the tangent line to the graph of a function at a particular point. Differentiation is an essential tool when dealing with curves and can help solve a variety of problems involving motion, optimization, and more.

When we differentiate a function, we are essentially looking for its derivative. The derivative, denoted as \( \frac{dy}{dx} \), represents the function's rate of change with respect to an independent variable, typically \( x \). For example, differentiating a simple function like \( y = x^2 \) gives us \( \frac{dy}{dx} = 2x \), indicating that the slope of the tangent to the curve \( y = x^2 \) changes linearly with \( x \).

In the exercise given, we're asked to differentiate the function \( y = \sinh(\cos 3x) \), a composition of functions, using the chain rule. Differentiation of composite functions like this requires careful attention to how each part of the function changes.
Hyperbolic Functions
Hyperbolic functions, analogous to trigonometric functions, are commonly used in calculus and engineering. They include functions such as \( \sinh(x) \) and \( \cosh(x) \), and are defined using exponential functions.

The hyperbolic sine function, \( \sinh(x) \), is defined as \( \sinh(x) = \frac{e^x - e^{-x}}{2} \), and it bears a strong resemblance to the sine function, but is not periodic. On the other hand, the hyperbolic cosine function, \( \cosh(x) \), is given by \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).

In differentiation, the derivatives of hyperbolic functions follow a simple pattern: the derivative of \( \sinh(x) \) is \( \cosh(x) \), and conversely, the derivative of \( \cosh(x) \) is \( \sinh(x) \).
  • \( \frac{d}{dx} \sinh(x) = \cosh(x) \)
  • \( \frac{d}{dx} \cosh(x) = \sinh(x) \)
These functions are particularly useful when dealing with problems involving hyperbolas, hence the name, and appear often in the descriptions of systems with exponential growth or decay.
Composition of Functions
The concept of composition of functions is central to understanding advanced calculus operations like the chain rule. When dealing with a function composed of two or more functions, we first need to identify the inner and outer functions.

Consider a composite function \( y = f(g(x)) \), where \( g(x) \) is the inner function and \( f(u) \) is the outer function, where \( u = g(x) \). In our exercise, \( \cos 3x \) is the inner function, and \( \sinh(u) \) is the outer function, where \( u = \cos 3x \).

To differentiate composite functions, the chain rule is used. The chain rule states that the derivative of \( y = f(g(x)) \) is \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
  • Differentiate the outer function with respect to the inner function: \( f'(u) \).
  • Differentiate the inner function with respect to \( x \): \( g'(x) \).
  • Multiply these derivatives to get the derivative of the composite function.
In our exercise, this involves differentiating \( \sinh(u) \) with respect to \( \cos 3x \), and then multiplying it by the derivative of \( \cos 3x \), resulting in the final expression for \( \frac{dy}{dx} \).

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Most popular questions from this chapter

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