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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 37

Calculate the derivative of the following functions.
y = sin (4x3 + 3x +1)

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1
Step 1: Identify the outer function and the inner function. Here, the outer function is \( \sin(u) \) and the inner function is \( u = 4x^3 + 3x + 1 \).
Step 2: Apply the chain rule for differentiation, which states that \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).
Step 3: Differentiate the outer function \( \sin(u) \) with respect to \( u \). The derivative is \( \cos(u) \).
Step 4: Differentiate the inner function \( u = 4x^3 + 3x + 1 \) with respect to \( x \). The derivative is \( 12x^2 + 3 \).
Step 5: Combine the results from steps 3 and 4 using the chain rule: \( \frac{dy}{dx} = \cos(4x^3 + 3x + 1) \cdot (12x^2 + 3) \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivative

The derivative of a function measures how the function's output value changes as its input value changes. It is a fundamental concept in calculus that provides the slope of the tangent line to the curve at any given point. The derivative is often denoted as f'(x) or dy/dx and can be calculated using various rules, such as the power rule, product rule, and chain rule.
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Chain Rule

The chain rule is a formula for computing the derivative of the composition of two or more functions. It states that if you have a function y = f(g(x)), the derivative is given by dy/dx = f'(g(x)) * g'(x). This rule is essential when differentiating functions that are nested within each other, such as the sine function in the given problem.
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Trigonometric Functions

Trigonometric functions, such as sine and cosine, relate angles to the ratios of sides in right triangles. In calculus, these functions are important because they have specific derivatives: for example, the derivative of sin(x) is cos(x). Understanding how to differentiate these functions is crucial when working with problems involving trigonometric expressions, as seen in the given function.
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