Problem 36 - Calculus Early Transcendentals

Question

Find \(d y / d x\) $$ y=\left(x^{2}+x\right)^{5} \sin ^{8} x $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = 5(x^2 + x)^4 (2x + 1) \sin^8 x + 8(x^2 + x)^5 \sin^7 x \cos x \).

Step 1: Identify the Rule of Differentiation

The function given is a product of two functions, \( (x^2 + x)^5 \) and \( \sin^8x \). Therefore, we will use the product rule to differentiate it.

Step 2: Apply the Product Rule

The product rule states that if you have a function \( y = u \cdot v \), then \( \frac{dy}{dx} = u'v + uv' \). Here \( u = (x^2 + x)^5 \) and \( v = \sin^8x \). We'll differentiate each part separately.

Step 3: Differentiate \( u = (x^2 + x)^5 \) Using the Chain Rule

Using the chain rule, let \( g(x) = x^2 + x \) so that \( u = g(x)^5 \). Then, \( u' = 5g(x)^4 g'(x) \). Since \( g(x) = x^2 + x \), we have \( g'(x) = 2x + 1 \). So, \( u' = 5(x^2 + x)^4 (2x + 1) \).

Step 4: Differentiate \( v = \sin^8x \) Using the Chain Rule

Again using the chain rule, let \( h(x) = \sin x \) so that \( v = h(x)^8 \). Then, \( v' = 8 h(x)^7 h'(x) \). Since \( h'(x) = \cos x \), \( v' = 8 \sin^7 x \cos x \).

Step 5: Compute \( \frac{dy}{dx} \) Using the Product Rule Results

Substitute \( u \), \( u' \), \( v \), and \( v' \) into the product rule: \[ \frac{dy}{dx} = 5(x^2 + x)^4 (2x + 1) \cdot \sin^8 x + (x^2 + x)^5 \cdot 8 \sin^7 x \cos x \].

Step 6: Simplify the Expression

Simplify the expression to make it more manageable if needed. As is, \( \frac{dy}{dx} = 5(x^2 + x)^4 (2x + 1) \sin^8 x + 8(x^2 + x)^5 \sin^7 x \cos x \) is already quite concise.

Key Concepts

Product Rule

To find the derivative of a product of two functions, we use the product rule. The product rule is a powerful tool in calculus, allowing us to deal with functions that are multiplied together. If you have a function defined as a product of two separate functions, say \( y = u \cdot v \), then the derivative \( \frac{dy}{dx} \) is given by the formula:

• \( \frac{dy}{dx} = u'v + uv' \)

Here, \( u' \) denotes the derivative of \( u \), and \( v' \) is the derivative of \( v \). This rule basically adds up the contributions of both functions to the derivative. The idea is to differentiate each part separately and then combine them. It ensures that the interaction between the multiplied functions is taken into account. This is crucial when dealing with more complex expressions involving products of functions.

Chain Rule

When functions are nested inside each other, like \((x^2 + x)^5\), we use the chain rule to differentiate them. The chain rule states that if a function \( y \) is composed of two functions \( g(x) \) and \( f(u) \), where \( u = g(x) \), then the derivative \( \frac{dy}{dx} \) is given by:

• \( \frac{dy}{dx} = f'(u) \cdot g'(x) \)

In simpler terms, differentiate the outer function and multiply it by the derivative of the inner function. For example, for \( (x^2 + x)^5 \), consider the inner function as \( g(x) = x^2 + x \) and the outer function as \( f(u) = u^5 \). The derivative then becomes \( 5(x^2 + x)^4 \cdot (2x + 1) \). This rule is especially helpful when dealing with power functions combined with other expressions, as well as composite functions like trigonometric functions with altered arguments.

Derivative of Trigonometric Functions

Differentiating trigonometric functions is a common task in calculus. These functions can be part of complex expressions, just like \( \sin^8 x \). Here, the chain rule aids in finding their derivative. To differentiate \( \sin^8x \), consider it as \( (\sin x)^8 \). Here, \( h(x) = \sin x \) serves as an inner function. The derivative involves:

• First differentiate the outer function: \( 8(\sin x)^7 \)
• Then multiply by the derivative of the inner function: \( \cos x \)

The result is: \( 8\sin^7x \cdot \cos x \). Trigonometric derivatives form the basis for more advanced calculus topics. Understanding how to differentiate these functions gives you the tools to tackle a wide range of problems. It's important to remember the basic trigonometric derivatives such as \( \frac{d}{dx} (\sin x) = \cos x \) and how to apply them in conjunction with the chain rule.