Problem 16 - 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. $$y=\left(1+x^{2}\right)\left(x^{3 / 4}-x^{-3}\right)$$

Step-by-Step Solution

Answer

Both methods result in: \(y' = \frac{3}{4}x^{-1/4} + \frac{11}{4}x^{7/4} + 3x^{-4} + x^{-2}\).

Step 1: Identify the Functions for Product Rule

The given function is \(y = (1 + x^2)(x^{3/4} - x^{-3})\). Identify \(u(x) = 1 + x^2\) and \(v(x) = x^{3/4} - x^{-3}\).

Step 2: Compute Derivatives

Differentiate each function: \(u'(x) = 2x\) and \(v'(x) = \frac{3}{4}x^{-1/4} + 3x^{-4}\).

Step 3: Apply the Product Rule

Apply the Product Rule: \(y' = u'v + uv'\). This yields \((2x)(x^{3/4} - x^{-3}) + (1 + x^2)(\frac{3}{4}x^{-1/4} + 3x^{-4})\).

Step 4: Simplify the Expression

Simplify the product: \(2x \cdot x^{3/4} - 2x \cdot x^{-3} + (1 + x^2)(\frac{3}{4}x^{-1/4}) + (1 + x^2)(3x^{-4})\). Then, simplify further to combine like terms.

Step 5: Expand the Original Expression

Multiply the factors directly: \((1 + x^2)(x^{3/4} - x^{-3}) = 1 \cdot x^{3/4} + x^2 \cdot x^{3/4} - 1 \cdot x^{-3} - x^2 \cdot x^{-3}\).

Step 6: Simplify the Expanded Expression

Combine the terms: \(x^{3/4} + x^{11/4} - x^{-3} - x^{-1}\).

Step 7: Differentiate Each Term

Differentiate: \((x^{3/4})' = \frac{3}{4}x^{-1/4}\), \((x^{11/4})' = \frac{11}{4}x^{7/4}\), \((x^{-3})' = -3x^{-4}\), \((x^{-1})' = x^{-2}\).

Step 8: Write Down the Final Result

Combine the differentiated terms of the expanded expression: \(y' = \frac{3}{4}x^{-1/4} + \frac{11}{4}x^{7/4} + 3x^{-4} + x^{-2}\).

Key Concepts

Product Rule

Differentiating a product of two functions requires the application of the product rule. This rule is central when dealing with derivatives involving products of two functions. To apply it, let's first identify the two functions we're dealing with. Assume we have a function in the form of \( y = u(x) \cdot v(x) \). The product rule tells us that the derivative \( y' \) is given by:

• \( y' = u'(x) \cdot v(x) + u(x) \cdot v'(x) \)

Here, \( u(x) \) and \( v(x) \) are functions of \( x \), and \( u'(x) \) and \( v'(x) \) are their respective derivatives. The challenge lies in correctly identifying and differentiating each function before plugging them back into the rule. This approach preserves the complexities of each function which are essential to fitting the puzzle together to find the derivative.

Simplification

Once you've applied the product rule, the next major step is simplification. Simplification is crucial for making the expression easy to handle and integrate further. It involves combining like terms and simplifying exponents. For complex expressions, break them down into smaller, manageable pieces.

Consider the expression

• \( 2x \cdot x^{3/4} - 2x \cdot x^{-3} + (1 + x^2)(\frac{3}{4}x^{-1/4} + 3x^{-4}) \)

Combine similar terms to consolidate parts of the expression. Focus on reducing the expression down to simpler terms, tirelessly looking for common factors or terms you can combine. Once you've streamlined your expression, you can often proceed with further differentiation or analysis.

Polynomial Differentiation

Polynomial differentiation involves finding the derivative of each term in a polynomial separately. This technique is straightforward and involves differentiating each term based on its own degree and coefficient.

• For a term \( x^n \), the derivative is \( nx^{n-1} \).
• Every term is treated as its own mini-problem, solved independently.

In our problem, the simplified expression contains terms like \( x^{3/4}, x^{11/4}, x^{-3}, \) and \( x^{-1} \). For each term:

• Differentiate using the power rule: \( n \cdot x^{n-1} \).
• Accumulate the results to form the derivative of the entire polynomial.

This approach reduces complex functions into manageable pieces that can be easily solved individually, leveraging the power of differentiating powers.

Exponentiation Rules

Exponent rules are key when working with differentiation, especially when dealing with powers of variables. These rules help in rewriting terms during processes like simplification or polynomial differentiation.

• \( a^m \cdot a^n = a^{m+n} \)
• \((a^m)^n = a^{mn} \)
• \( a^m / a^n = a^{m-n} \)
• Negative exponents indicate reciprocal: \( a^{-n} = 1/a^n \)

These rules help transform the expressions involved in differentiation to maintain consistency in terms and to manage expressions fluidly. For example, multiplying expressions like \( x^2 \cdot x^{3/4} \) results in \( x^{2 + 3/4} = x^{11/4} \). Applying these rules is necessary for both accurate simplification and differentiation.