Chapter 3: Problem 44
Find \(d y / d x\). \(y=\sin \sqrt{3 x+1}\)
Short Answer
Expert verified
\( \frac{dy}{dx} = \frac{3\cos(\sqrt{3x + 1})}{2\sqrt{3x + 1}} \)
Step by step solution
01
Identify the Inner Function
The given function is in the form of a composite function, where the outer function is a sine function and the inner function is a square root function, \( u = \sqrt{3x + 1} \). So, you should identify \( u \) such that \( y = \sin(u) \) with \( u = \sqrt{3x + 1} \).
02
Differentiate the Outer Function
Differentiate the outer function with respect to \( u \). The derivative \( \frac{d}{du}(\sin(u)) = \cos(u) \).
03
Differentiate the Inner Function
Differentiate the inner function \( u = \sqrt{3x + 1} \) with respect to \( x \). Use the chain rule for the square root: \( \frac{d}{dx}(\sqrt{3x + 1}) = \frac{1}{2\sqrt{3x + 1}} \times 3 = \frac{3}{2\sqrt{3x + 1}} \).
04
Apply the Chain Rule
To find \( \frac{dy}{dx} \), apply the chain rule: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \cos(\sqrt{3x + 1}) \cdot \frac{3}{2\sqrt{3x + 1}} \).
05
Simplify the Result
The derivative \( \frac{dy}{dx} = \frac{3\cos(\sqrt{3x + 1})}{2\sqrt{3x + 1}} \). Ensure there are no common terms to simplify further, and this is your final result.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Chain Rule
The chain rule is a fundamental technique in calculus used to differentiate composite functions. It allows us to take the derivative of a function composed of two or more functions. For instance, if you have a function expressed as \( y = f(g(x)) \), the chain rule states that the derivative of \( y \) with respect to \( x \) is the product of the derivative of \( f \) with respect to \( g \), and the derivative of \( g \) with respect to \( x \). This can be mathematically represented as:
So, when you look at functions that are nested, like a sine function inside a square root, the chain rule efficiently manages the complexity. The key is to break it down into parts:
- \( \frac{dy}{dx} = \frac{df}{dg} \times \frac{dg}{dx} \)
So, when you look at functions that are nested, like a sine function inside a square root, the chain rule efficiently manages the complexity. The key is to break it down into parts:
- Differentiate the outer function first.
- Then, multiply by the derivative of the inner function.
Decomposing Composite Functions
Composite functions occur when one function is nested inside another. For example, in the function \( y = \sin(\sqrt{3x + 1}) \), the sine function contains a square root function as its argument. To differentiate such a function, it's crucial to recognize the 'inner' and 'outer' components. Here:
- The outer function is \( \sin(u) \).
- The inner function is \( u = \sqrt{3x + 1} \).
Mastery of Trigonometric Differentiation
Trigonometric differentiation is vital because trigonometric functions frequently appear in many areas of calculus. Derivatives of trigonometric functions follow specific rules:
When these functions are part of a composite function, like \( \sin(\sqrt{3x + 1}) \), these basic rules blend with the chain rule. Begin by acknowledging the outermost trigonometric function and use its differentiation rule. Then, through the chain rule, continue differentiating any inner non-trigonometric parts. The composite nature means you multiply the derivatives together, forming a bridge from the general rules of differentiation to specific trigonometric applications. With practice, you'll handle both trigonometric and composite functions simultaneously with accuracy and ease.
- The derivative of \( \sin(x) \) is \( \cos(x) \).
- The derivative of \( \cos(x) \) is \( -\sin(x) \).
- For \( \tan(x) \), the derivative is \( \sec^2(x) \).
When these functions are part of a composite function, like \( \sin(\sqrt{3x + 1}) \), these basic rules blend with the chain rule. Begin by acknowledging the outermost trigonometric function and use its differentiation rule. Then, through the chain rule, continue differentiating any inner non-trigonometric parts. The composite nature means you multiply the derivatives together, forming a bridge from the general rules of differentiation to specific trigonometric applications. With practice, you'll handle both trigonometric and composite functions simultaneously with accuracy and ease.