Chapter 27: Problem 5
Find the derivatives of the given functions. $$y=2 \sin \left(2 x^{3}-1\right)$$
Short Answer
Expert verified
The derivative is \( \frac{dy}{dx} = 12x^2 \cos(2x^3 - 1) \).
Step by step solution
01
Identify the Outer and Inner Functions
The function given is a composition of two functions: the outer function is the sine function, and the inner function is the expression inside the sine. Let the outer function be \( f(u) = 2 \sin(u) \) where \( u = 2x^3 - 1 \).
02
Find the Derivative of the Outer Function
The derivative of the outer function \( f(u) = 2 \sin(u) \) with respect to \( u \) is \( f'(u) = 2 \cos(u) \). This means the derivative of the sine component is scaled by a factor of 2.
03
Find the Derivative of the Inner Function
The inner function is \( u = 2x^3 - 1 \). The derivative of this function with respect to \( x \) is \( \frac{du}{dx} = 6x^2 \).
04
Apply the Chain Rule
The derivative of the original function \( y = 2 \sin(2x^3 - 1) \) can be found by applying the chain rule, which states \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). Therefore, \( \frac{dy}{dx} = 2 \cos(2x^3 - 1) \cdot 6x^2 \).
05
Simplify the Expression
Simplify the product of derivatives: \( \frac{dy}{dx} = 12x^2 \cos(2x^3 - 1) \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
In calculus, derivatives play a crucial role in understanding how functions change. The derivative of a function gives us the rate at which the function's value changes with respect to its input. It is a foundational concept in calculus that provides insights into the behavior and properties of functions.
There are several key points to understand about derivatives:
There are several key points to understand about derivatives:
- They indicate the slope of a function at any given point.
- They help find maxima and minima of functions, which are the highest or lowest points.
- They are used to determine the instantaneous rate of change, similar to how speed measures the rate of change of distance with time.
Chain Rule
The chain rule is a fundamental tool in calculus for finding the derivative of composite functions. A composite function is a function made up of other functions. This rule allows us to differentiate functions that are nested inside one another.
To apply the chain rule, follow these steps:
To apply the chain rule, follow these steps:
- Identify the inner and outer functions within the composite function. In our case, the outer function is the sine function, and the inner function is the polynomial inside the sine.
- Differentially the outer function with respect to the inner function, treating the entire inner function as a single variable.
- Differentiate the inner function with respect to its variable.
- Multiply these derivatives together to get the derivative of the original function.
Trigonometric Functions
Trigonometric functions like sine, cosine, and tangent are essential in calculus, frequently appearing in many real-world problems and mathematical equations. These functions model periodic phenomena, such as waves and oscillations.
For derivatives involving trigonometric functions, it's crucial to understand their unique properties:
For derivatives involving trigonometric functions, it's crucial to understand their unique properties:
- The derivative of \( \sin(x) \) is \( \cos(x) \).
- The derivative of \( \cos(x) \) is \( -\sin(x) \).
- Scaling these functions, such as in our outer function \( 2\sin(u) \), affects the magnitude but not the form. The rules still apply.