Problem 216 - Calculus Volume 1

Question

For the following exercises, given \(y=f(u) \quad\) and \(u=g(x),\) find \(\frac{d y}{d x}\) by using Leibniz's notation for the chain rule: \(\frac{d y}{d x}=\frac{d y}{d u} \frac{d u}{d x}\) $$y=\sin u, u=5 x-1$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = 5 \cos(5x - 1) \)

Step 1: Identify the Functions

We are given that \( y = \sin u \) and \( u = 5x - 1 \). Here, \( y \) is a function of \( u \), and \( u \) is a function of \( x \). We need to find the derivative of \( y \) with respect to \( x \) using Leibniz's notation for the chain rule.

Step 2: Differentiate \( y \) with Respect to \( u \)

To find \( \frac{dy}{du} \), we differentiate \( y = \sin u \) with respect to \( u \). The derivative of \( \sin u \) is \( \cos u \), therefore:\[ \frac{dy}{du} = \cos u \]

Step 3: Differentiate \( u \) with Respect to \( x \)

To find \( \frac{du}{dx} \), differentiate \( u = 5x - 1 \) with respect to \( x \). The derivative of \( 5x - 1 \) is 5, thus:\[ \frac{du}{dx} = 5 \]

Step 4: Apply the Chain Rule

Now apply the chain rule formula for derivatives in Leibniz's notation:\[ \frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx} = (\cos u)(5) \]This simplifies to:\[ \frac{dy}{dx} = 5 \cos u \]

Step 5: Substitute \( u \) Back in Terms of \( x \)

Substitute \( u = 5x - 1 \) back into the expression for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = 5 \cos(5x - 1) \]

Key Concepts

Leibniz's Notation

Leibniz's notation is a widely used method for writing derivatives, named after the mathematician Gottfried Wilhelm Leibniz. It's expressed in the form \( \frac{dy}{dx} \), indicating the derivative of \( y \) with respect to \( x \). This notation is extremely helpful when dealing with composite functions, where one function is nested inside another.

In the given exercise, Leibniz's notation is utilized to express the chain rule. The chain rule is a formula for computing the derivative of a composite function. When you see \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \), it simply means that to find the derivative of \( y \) concerning \( x \), we identify the rate of change of \( y \) with respect to \( u \) and multiply it with the rate of change of \( u \) with respect to \( x \).

With this notation, it's easier to visualize how each segment of a composite function affects the derivative of the overall function. This clarity makes it ideal for communicating complex calculus concepts efficiently.

Differentiation

Differentiation is the process of finding the derivative of a function. It translates how a function's value changes as its inputs change, essentially capturing the function's rate of change. In the exercise, the task is to differentiate nested functions using the chain rule.

The differentiation of \( y = \sin u \) with respect to \( u \) gives us \( \frac{dy}{du} = \cos u \). This step is about applying basic trigonometric differentiation, where the derivative of the sine function is cosine. Next, for the outer function, \( u = 5x - 1 \), differentiating with respect to \( x \) yields \( \frac{du}{dx} = 5 \). This shows that for linear functions like \( 5x-1 \), the derivative is a constant equal to the coefficient of \( x \).

By differentiating both components separately, we prepare the ground to apply the chain rule to find the derivative of \( y \) concerning \( x \), which results in \( 5\cos(5x - 1) \).

Sine Function

The sine function, often denoted as \( \sin \), is a fundamental trigonometric function. It is periodic and relates to the angles of a right triangle and the coordinates of a point on a unit circle. In calculus, the sine function's derivative is always a cosine function.

In this exercise, \( y = \sin u \) implies that the sine function is in play as part of a composite scenario. Differentiating this gives \( \frac{dy}{du} = \cos u \), meaning the rate of change of \( y \) when \( u \) changes is represented by the cosine of \( u \).

The transformation from sine to cosine during differentiation is rooted in the geometric properties of the unit circle, where the derivative at any point is reflected by the adjacent side as it moves along the circle. This periodic relationship is key when involving the sine function in calculus.

Composite Functions

A composite function is formed when one function is substituted into another, represented as \( f(g(x)) \). In the exercise, \( y = \sin(5x - 1) \) represents a composite function where \( \sin u \) and \( u = 5x - 1 \) combine. Handling such functions requires the chain rule for differentiation.

To differentiate a composite function, find the derivative of the outer function first while keeping the inner function intact. For instance, \( \frac{dy}{du} \) gives the change of \( y \) concerning \( u \) in the outer layer. Then, calculate \( \frac{du}{dx} \), determining how \( u \) changes with \( x \). By multiplying these derivatives, you leverage the chain rule: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).

The result \( 5\cos(5x - 1) \) marks the derivative of the entire composite function, showing how an alteration in \( x \) impacts \( y \) by simultaneously considering changes in both the inner and outer functions.