Problem 36 - Calculus - AP Edition

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Function Components

The given function is \( y = \left(x^2 + x\right)^5 \sin^8 x \). We notice that it is a product of two functions: \( u = (x^2 + x)^5 \) and \( v = \sin^8 x \).

Step 2: Apply the Product Rule

The product rule states that if \( y = u \, v \), then \( \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \). We must find the derivatives of \( u \) and \( v \) separately.

Step 3: Differentiate the First Function

To find \( \frac{du}{dx} \), we use the chain rule. Let \( u = (x^2 + x)^5 \). Differentiating, we have \( \frac{du}{dx} = 5(x^2 + x)^4 \cdot \frac{d}{dx}(x^2 + x) \). The derivative of \( x^2 + x \) is \( 2x + 1 \). Thus, \( \frac{du}{dx} = 5(x^2 + x)^4 (2x + 1) \).

Step 4: Differentiate the Second Function

Now, find \( \frac{dv}{dx} \) where \( v = \sin^8 x \). Using the chain rule, \( \frac{dv}{dx} = 8 \sin^7 x \cdot \cos x \).

Step 5: Substitute into the Product Rule Formula

Substitute the results from steps 3 and 4 into the product rule formula: \( \frac{dy}{dx} = (x^2 + x)^5 [8 \sin^7 x \cdot \cos x] + \sin^8 x [5(x^2 + x)^4 (2x + 1)] \).

Step 6: Simplify the Expression

Combining terms, the derivative is: \( \frac{dy}{dx} = 8 (x^2 + x)^5 \sin^7 x \cos x + 5 \sin^8 x (x^2 + x)^4 (2x + 1) \). This is the simplified expression for \( \frac{dy}{dx} \).

Key Concepts

Product Rule

The product rule is a fundamental tool in calculus used for differentiating products of two functions. It is essential when dealing with expressions where functions are multiplied together, just like in the exercise problem. If you have a function, say \( y = u(x) \, v(x) \), the product rule states that the derivative of \( y \) with respect to \( x \) is \((u \, v)' = u \frac{dv}{dx} + v \frac{du}{dx}\).

Here's how you can remember it:

• Differentiate the first function, \( u \), while leaving the second function, \( v \), unchanged.
• Then, differentiate the second function, \( v \), while leaving the first function, \( u \), unchanged.
• Finally, sum these two results.

This process ensures no component of the product is neglected when calculating the derivative. In our example exercise, the functions \( u = (x^2 + x)^5 \) and \( v = \sin^8 x \) were identified and differentiated separately before applying the formula.

Chain Rule

The chain rule is another vital differentiation technique used when a function is composed of other functions. It's particularly useful for finding derivatives of composite functions, where one function is nested inside another.

The chain rule states: If a variable \( z \) depends on the variable \( y \), which in turn depends on the variable \( x \), then \( \frac{dz}{dx} = \frac{dz}{dy} \cdot \frac{dy}{dx} \).

In simpler terms:

• Differentiating an outside function and leaving the inside function unchanged.
• Then, multiply by the derivative of the inside function.

In the exercise solution, the chain rule was applied when differentiating \( u = (x^2 + x)^5 \) and \( v = \sin^8 x \):

• For \( u \), it led to \( 5(x^2 + x)^4 \cdot (2x + 1) \).
• For \( v \), it led to \( 8 \sin^7 x \cdot \cos x \).

The result combined the outer derivative with the derivative of the inside function.

Differentiation Techniques

Differentiation techniques like the product rule and chain rule are handy tools when solving calculus problems involving derivatives. Understanding their applications can ease the task of finding derivatives of complex functions.

Each of these techniques has its placement in particular situations. If a function is a product of two separable functions, the product rule is the go-to technique. Conversely, if you're dealing with a function that includes another function's composition, the chain rule is fitting.

Stacking these methods allows for tackling intricate problems by breaking them down into simpler steps. In many practical exercises, identifying which technique to use and in what sequence can significantly simplify the calculus process.

The goal of these rules is to make calculus manageable and systematic, ensuring that no detail is missed in determining the rate of change or slope of a function at any given point.