Problem 222 - Calculus Volume 1

Question

For the following exercise, a. decompose each function in the form \(y=f(u)\) and \(u=g(x),\) and b. find \(\frac{d y}{d x}\) as a function of \(x\). $$ y=\sin ^{5}(x) $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = 5(\sin(x))^4 \cdot \cos(x) \).

Step 1: Identify Inner Function

We begin by identifying the inner function, which is the part of the function inside the outer function. The given function is \( y = \sin^5(x) \). Here, \( \sin(x) \) serves as the inner function. Let \( u = g(x) = \sin(x) \).

Step 2: Identify Outer Function

Next, we identify the outer function. The entire expression \( \sin^5(x) \) indicates that the outer function is applying a power of 5. Therefore, we can express the outer function as \( y = f(u) = u^5 \).

Step 3: Differentiate Outer Function

Now, differentiate the outer function \( y = f(u) = u^5 \) with respect to \( u \). This results in \( \frac{dy}{du} = 5u^4 \).

Step 4: Differentiate Inner Function

Differentiate the inner function \( u = \sin(x) \) with respect to \( x \), resulting in \( \frac{du}{dx} = \cos(x) \).

Step 5: Apply Chain Rule

Apply the chain rule to find \( \frac{dy}{dx} \). The chain rule states that \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). Substituting the derivatives, we get \( \frac{dy}{dx} = 5u^4 \cdot \cos(x) \).

Step 6: Substitute Back Inner Function

Replace \( u \) back with \( \sin(x) \) for the final solution: \( \frac{dy}{dx} = 5(\sin(x))^4 \cdot \cos(x) \).

Key Concepts

Chain Rule

The chain rule is a powerful tool in calculus used for differentiating compositions of functions. When you have a function that is composed of two or more functions, the chain rule helps to find the derivative of this composite function. Think of it as a way to "chain" the derivatives together by calculating the derivative of the outer function and multiplying it by the derivative of the inner function.
To use the chain rule, remember this formula:

• \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)

This means that you first differentiate the outer function with respect to its variable and then multiply it by the derivative of the inner function with respect to \(x\).
In the original exercise, the outer function is \(u^5\), and the inner function is \(\sin(x)\). So, we apply the chain rule by differentiating \(u^5\) to get \(5u^4\) and \(\sin(x)\) to get \(\cos(x)\), then multiply these two to find the final derivative.

Composite Functions

Composite functions are functions made up of two or more simpler functions. For instance, in the expression \(y = \sin^5(x)\), we can see it's a composite: the \(\sin(x)\) function is inside the power function. In more formal terms, let

• \( y = f(u) \)
• \( u = g(x) \)

The function \( f(g(x)) \) is a composite function.
The decomposition of these functions into their individual components is a helpful step when applying the chain rule. By identifying the inner and outer functions, we better understand how to apply calculus operations like differentiation. In our exercise, the inner part \(\sin(x)\) is transformed by the outer function's operation of taking it to the 5th power.

Derivatives

Derivatives are foundational to calculus, representing how a function changes as its input changes. They are visualized as slopes of tangent lines to the function's graph.
In differentiation, several rules apply. The power rule, often used here, tells us how to differentiate functions of the form \(u^n\), resulting in \(nu^{n-1}\). In the context of our exercise:

• The derivative of \(u^5\) with respect to \(u\) is \(5u^4\).
• The derivative of \(\sin(x)\) with respect to \(x\) is \(\cos(x)\).

By taking these derivatives and utilizing the chain rule, we found the derivative of the composite function \(\sin^5(x)\) effectively. Remember, derivatives tell us about the function's behavior and growth at each point along its curve.