Problem 27 - Calculus Early Transcendentals

Question

Find \(d y / d x\) $$ y=x^{3} \sin ^{2}(5 x) $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = 3x^2 \sin^2(5x) + 5x^3 \sin(10x) \).

Step 1: Identify Function Components

The given function is \( y = x^3 \sin^2(5x) \). To differentiate this, notice that it is a product of \( x^3 \) and \( \sin^2(5x) \). To find \( \frac{dy}{dx} \), you will apply the product rule.

Step 2: Apply Product Rule

The product rule states that if \( y = u \cdot v \), then \( \frac{dy}{dx} = u'v + uv' \). Here, let \( u = x^3 \) and \( v = \sin^2(5x) \). Find the derivatives \( u' \) and \( v' \) separately.

Step 3: Differentiate \( u = x^3 \)

Differentiate \( u = x^3 \) with respect to \( x \):\[\frac{du}{dx} = 3x^2.\]

Step 4: Differentiate \( v = \sin^2(5x) \)

To differentiate \( v = \sin^2(5x) \), use the chain rule. Let \( w = \sin(5x) \), so \( v = w^2 \). By the chain rule:\[\frac{dv}{dx} = 2w \cdot \frac{dw}{dx}.\]First, differentiate \( w = \sin(5x) \), which gives:\[\frac{dw}{dx} = 5\cos(5x).\]Thus:\[\frac{dv}{dx} = 2 \sin(5x) \cdot 5 \cos(5x) = 10 \sin(5x) \cos(5x).\]Recognize that \( 2 \sin(a) \cos(a) = \sin(2a) \), so:\[\frac{dv}{dx} = 5 \sin(10x).\]

Step 5: Combine Using Product Rule

Substitute \( u \), \( v \), and their derivatives back into the product rule:\[\frac{dy}{dx} = (3x^2)(\sin^2(5x)) + (x^3)(5\sin(10x)).\]

Step 6: Simplify the Expression

Distribute and simplify the result from the previous step:\[\frac{dy}{dx} = 3x^2 \sin^2(5x) + 5x^3 \sin(10x).\]

Step 7: Final Answer

The derivative \( \frac{dy}{dx} \) is:\[\frac{dy}{dx} = 3x^2 \sin^2(5x) + 5x^3 \sin(10x).\]

Key Concepts

Product Rule

The product rule is a fundamental concept in calculus used to differentiate the product of two functions. If you have two functions, say, \( u(x) \) and \( v(x) \), their product is \( y(x) = u(x) \cdot v(x) \). The product rule tells us how to find the derivative of this product: \[ \frac{dy}{dx} = u'(x)\cdot v(x) + u(x)\cdot v'(x). \] This means that to find the derivative of a product, you differentiate each function separately and then apply this formula. In our original exercise, we identified the function as \( y = x^3 \cdot \sin^2(5x) \). Here, \( u(x) \) is \( x^3 \) and \( v(x) \) is \( \sin^2(5x) \).

• First, differentiate \( u(x) = x^3 \): The derivative \( u'(x) \) is \( 3x^2 \).
• Next, differentiate \( v(x) = \sin^2(5x) \): We will use the chain rule for this step, which we will discuss next.

Finally, we combined these results using the product rule to find the derivative of the entire function. The product rule is essential especially for functions combining powers with trigonometric expressions or any two differentiable components.

Chain Rule

The chain rule is another pivotal tool in calculus, particularly useful for differentiating compositions of functions. If you have a function \( y = f(g(x)) \), the chain rule states that: \[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x). \] Essentially, you differentiate the outer function and multiply it by the derivative of the inner function. In the exercise, for the term \( \sin^2(5x) \), we noticed that there is a composition involving the square function and a trigonometric function inside it.

• Recognize the hierarchy: The outer function here is \( u(t) = t^2 \), where \( t = \sin(5x) \).
• Apply the chain rule: Differentiate the outer function first, which yields \( 2t \cdot t'(x) \) or \( 2 \sin(5x) \cdot \frac{d}{dx}(\sin(5x)) \).
• This means: the derivative \( \frac{dw}{dx} = 5\cos(5x) \). Thus, \( 10\sin(5x)\cos(5x) \) simplifies to \( 5\sin(10x) \) using the double-angle formula for sine.

The chain rule played a crucial role in unpacking the derivative of the nested trigonometric functions within the square. It helps break down complicated functions into simpler parts, making the differentiation process more manageable.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent often appear in calculus problems due to their periodic properties. Understanding how to differentiate these functions is crucial for success in calculus.

• Basic Derivatives: Remember, the derivative of \( \sin(x) \) is \( \cos(x) \), and the derivative of \( \cos(x) \) is \(-\sin(x) \). These are essential relationships to memorize.
• Function Composition: When trigonometric functions apply to more complicated inner functions, as in our exercise \( \sin^2(5x) \), we use the chain rule to differentiate the inner function first (\( 5x \), for example, as its derivative is \( 5 \)) and then apply the basic trigonometric derivative rules.
• Utilize Identities: Often, trigonometric identities can simplify your work. For instance, the formula \( \sin(2a) = 2\sin(a)\cos(a) \) helped simplify \( 10\sin(5x)\cos(5x) \) to \( 5\sin(10x) \).

Sine and cosine functions are the backbone for deeper trigonometric analysis and modeling periodic behavior. Mastery of these functions and their derivatives facilitates not only solving exercises but also understanding their application in physical models.