Problem 3 - Thomas Calculus

Question

In Exercises \(1-8,\) given \(y=f(u)\) and \(u=g(x),\) find \(d y / d x=\) \(d y / d x=f^{\prime}(g(x)) g^{\prime}(x)\) $$ y=\sin u, \quad u=3 x+1 $$

Step-by-Step Solution

Answer

The derivative \(dy/dx = 3\cos(3x + 1).\)

Step 1: Differentiate the outer function

The function given is composed of two functions: an outer function \(y = \sin(u)\), and an inner function \(u = 3x + 1\). To find the derivative of the outer function \(y = \sin(u)\) with respect to its argument \(u\), we use \(f'(u) = \cos(u)\).

Step 2: Differentiate the inner function

The inner function is \(u = g(x) = 3x + 1\). Differentiate this function with respect to \(x\) to get \(g'(x) = 3\).

Step 3: Apply the chain rule

The chain rule states that \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\). Substitute \(f'(u) = \cos(u)\) and \(g'(x) = 3\) into the chain rule to compute: \(\frac{dy}{dx} = \cos(3x + 1) \times 3\).

Step 4: Simplify the derivative

Simplify the expression obtained from the chain rule to get the final solution: \(\frac{dy}{dx} = 3\cos(3x + 1)\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, which deals with how a function changes as its input changes. Specifically, it is the process of finding the derivative. The derivative provides the slope of the function at any point on its curve, offering insight into the rate at which something is changing.

• To differentiate a function, you apply different rules depending on its form. Basic rules include power, product, quotient, and chain rules.
• In this exercise, differentiation is applied through the chain rule, a method used when dealing with compositions of functions.

When differentiating a composite function like the one in our exercise, it's essential to identify the inner and outer functions. Understanding this helps us apply the rules more effectively, enhancing our ability to solve calculus problems.

Composite Functions

Composite functions involve combining two or more functions, where the output of one function becomes the input of the next. In mathematical notation, a composite function is usually expressed as \(f(g(x))\) where \(g(x)\) is the inner function and \(f(u)\) is the outer function.

• The inner function \(g(x)\) is substituted into the outer function \(f(u) = \sin(u)\).
• The composite nature allows for complex operations on functions by applying each successively.

The complexity of composite functions lies in the fact that each layer of function affects the final derivative outcome. Recognizing which function is internal and which is external is crucial in applying the chain rule accurately.

Trigonometric Functions

Trigonometric functions are special kinds of functions in mathematics that relate angles to side lengths in right-angled triangles. The sine function, denoted as \(\sin(u)\), is one of the basic trigonometric functions.

• The sine function relates the angle of a triangle to the ratio of the length of the opposite side over the hypotenuse.
• In calculus, the derivative of the sine function is the cosine function, \(\cos(u)\).

In our exercise, \(\sin(u)\) serves as the outer function. Understanding how trigonometric functions differentiate allows us to effectively solve problems involving angular relationships and periodic behaviors in calculus. Mastery of these functions is essential, as they frequently appear in various areas of science and engineering.