Problem 14 - University Calculus: Early Transcendentals

Question

Find \(y^{\prime}\) (a) by applying the Product Rule and (b) by multiplying the factors to produce a sum of simpler terms to differentiate. $$y=(2 x+3)\left(5 x^{2}-4 x\right)$$

Step-by-Step Solution

Answer

Using both methods, the derivative is \( y' = 30x^2 + 22x - 12 \).

Step 1: Identifying the Functions for the Product Rule

Given the function \( y = (2x + 3)(5x^2 - 4x) \), we identify two functions: \( u = 2x + 3 \) and \( v = 5x^2 - 4x \). These will be used in the Product Rule for differentiation.

Step 2: Differentiating Each Function Separately

Next we need to find the derivatives of \( u \) and \( v \).For \( u = 2x + 3 \), the derivative \( u' = 2 \).For \( v = 5x^2 - 4x \), the derivative \( v' = 10x - 4 \).

Step 3: Applying the Product Rule

The Product Rule states \( (uv)' = u'v + uv' \).Substituting the derivatives: \( y' = (2)(5x^2 - 4x) + (2x + 3)(10x - 4) \).

Step 4: Simplifying the Expression from the Product Rule

Simplify each term: \( 2(5x^2 - 4x) = 10x^2 - 8x \) \( (2x + 3)(10x - 4) = 20x^2 - 8x + 30x - 12 \)Combine like terms: \( 10x^2 - 8x + 20x^2 + 30x - 12 \).

Step 5: Combining and Simplifying Final Derivative

Combine and simplify: \( 10x^2 + 20x^2 = 30x^2 \) and \(-8x + 30x = 22x \). The final derivative is \( y' = 30x^2 + 22x - 12 \).

Step 6: Expanding the Original Function

Expand \( y = (2x + 3)(5x^2 - 4x) \) using distribution: \( y = 2x(5x^2 - 4x) + 3(5x^2 - 4x) \).This gives \( 10x^3 - 8x^2 + 15x^2 - 12x \).

Step 7: Simplifying Expanded Expression

Combine like terms in the expanded expression: \( 10x^3 + 7x^2 - 12x \).

Step 8: Differentiating the Expanded Expression

Differentiate the simpler polynomial \( y = 10x^3 + 7x^2 - 12x \) term by term: \((10x^3)' = 30x^2\), \((7x^2)' = 14x\), \((12x)' = 12\).Thus, the derivative \( y' = 30x^2 + 14x - 12 \).

Key Concepts

Product Rule

To understand the product rule, we first need to break down the function given in the exercise. The product rule is a fundamental method in calculus used to differentiate products of two functions. The rule is stated as follows: if you have two differentiable functions, say \(u(x)\) and \(v(x)\), their derivative when multiplied together is given by \((uv)' = u'v + uv'\). So, you need to:

• Differentiate each function separately.
• Multiply the derivative of the first function by the second function (in its original form).
• Multiply the first function (in its original form) by the derivative of the second function.
• Add these two results together to get the derivative of the product.

From the original exercise, the two functions are \(u = 2x + 3\) and \(v = 5x^2 - 4x\). Following the product rule, differentiate \(u(x)\) to get \(u' = 2\), and \(v(x)\) to get \(v' = 10x - 4\). Here, you see:

• \(u'v = 2(5x^2 - 4x) = 10x^2 - 8x\)
• \(uv' = (2x + 3)(10x - 4) = 20x^2 + 30x - 8x - 12\)

After combining like terms, you get the simplified form \(30x^2 + 22x - 12\), which gives you a clear understanding of how differentiation works using the product rule.

Polynomial Expansion

When dealing with complex expressions, expanding polynomials can significantly simplify the process of differentiation. Polynomial expansion involves distributing and combining all like terms until the expression is written as a sum of simple polynomials. In the example given, you have the expression \((2x + 3)(5x^2 - 4x)\) which can be expanded by distributing as:

• \(2x \cdot 5x^2 = 10x^3\)
• \(2x \cdot (-4x) = -8x^2\)
• \(3 \cdot 5x^2 = 15x^2\)
• \(3 \cdot (-4x) = -12x\)

Combine these results to get \(10x^3 + 7x^2 - 12x\). Writing it this way eliminates the need to apply the product rule directly and provides a polynomial that is much easier to differentiate using basic rules of differentiation for monomials.

Derivatives

Understanding derivatives is essential in calculus, as they represent the rate at which a function is changing at any given point. Derivatives are calculated using different techniques, depending on the form of the function. After expanding a polynomial, each term can be differentiated separately due to the power rule which states that for a monomial \(ax^n\), its derivative is \(anx^{n-1}\).
In the example, after expansion, the polynomial \(10x^3 + 7x^2 - 12x\) was obtained. Let's differentiate each term:

• The derivative of \(10x^3\) is given by \(30x^2\), applying the power rule \(3 \cdot 10x^{3-1} = 30x^2\).
• The derivative of \(7x^2\) is \(14x\), using the same rule \(2 \cdot 7x^{2-1} = 14x\).
• The derivative of \(-12x\) is \(-12\), as the power rule simplifies to the coefficient when the exponent is 1.

Putting these together, you get \(30x^2 + 14x - 12\) which matches the derivative when using the product rule, showing the consistency and versatility in the methods used to find derivatives.