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In Exercises 13-16, find \(y^{\prime}\) (a) by applying the Product Rule and (b) by multiplying the factors to produce a sum of simpler terms to differentiate. \(y=\left(1+x^{2}\right)\left(x^{3 / 4}-x^{-3}\right)\)

Short Answer

Expert verified
The derivative is \( y' = \frac{11}{4}x^{7/4} + \frac{3}{4}x^{-1/4} + x^{-2} - 3x^{-4} \).

Step by step solution

01

Apply the Product Rule

The Product Rule states that if you have two functions \( u(x) \) and \( v(x) \), the derivative of their product \( u(x)v(x) \) is \( u'(x)v(x) + u(x)v'(x) \). Define \( u(x) = 1 + x^2 \) and \( v(x) = x^{3/4} - x^{-3} \). Differentiate each function separately:\[ u'(x) = \frac{d}{dx}(1 + x^2) = 2x \]\[ v'(x) = \frac{d}{dx}(x^{3/4} - x^{-3}) = \frac{3}{4}x^{-1/4} + 3x^{-4} \]Now apply the Product Rule:\[ y' = u'(x)v(x) + u(x)v'(x) = 2x(x^{3/4} - x^{-3}) + (1 + x^2)(\frac{3}{4}x^{-1/4} + 3x^{-4}) \]
02

Simplify the Expression

Now simplify the expression:\[ y' = 2x \cdot x^{3/4} - 2x \cdot x^{-3} + (1 + x^2)(\frac{3}{4}x^{-1/4}) + (1 + x^2)(3x^{-4}) \]Distributing and combining like terms:\[ y' = 2x^{7/4} - 2x^{-2} + \frac{3}{4}x^{-1/4} + \frac{3}{4}x^{7/4} + 3x^{-4} + 3x^{-2} \]Combine terms:\[ y' = (2 + \frac{3}{4})x^{7/4} + (-2 + 3)x^{-2} + \frac{3}{4}x^{-1/4} + 3x^{-4} \]\[ y' = \frac{11}{4}x^{7/4} + x^{-2} + \frac{3}{4}x^{-1/4} + 3x^{-4} \]
03

Expand the Original Expression (Alternative Method)

First, expand the original function by multiplying the binomials:\[ y = (1 + x^2)(x^{3/4} - x^{-3}) = 1 \cdot x^{3/4} + x^2 \cdot x^{3/4} - 1 \cdot x^{-3} - x^2 \cdot x^{-3} \]Calculate each term:\[ y = x^{3/4} + x^{11/4} - x^{-3} - x^{-1} \]This gives us the expanded form of \( y \) such that:\[ y = x^{11/4} + x^{3/4} - x^{-3} - x^{-1} \]
04

Differentiate the Expanded Form

Now differentiate this expanded expression term-by-term:\[ y' = \frac{11}{4}x^{7/4} + \frac{3}{4}x^{-1/4} - 3x^{-4} + x^{-2} \]This matches the result from following the Product Rule method.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule
The Product Rule is an essential technique in Calculus for finding the derivative of a product of two functions. This rule is indispensable when you have expressions where two functions are multiplied together. Let's break it down simply.
The Product Rule states: if you have two functions, say \( u(x) \) and \( v(x) \), the derivative of their product is given by:
  • \( (uv)' = u'v + uv' \)
This means that you differentiate the first function \( u(x) \) and multiply it by \( v(x) \), then add the product of \( u(x) \) with the derivative of \( v(x) \). It's a bit like saying: "differentiate the first, leave the second; differentiate the second, leave the first, then add.鈥
In the not-so-difficult task at hand, you consider each of your terms separately. For quick recall, remember: the order of differentiation doesn鈥檛 matter as long as you follow this logical application of the rule. Once you apply this correctly, cleaning up the terms is the key to getting simplified results. Should multiplication first look intimidating, try applying the Product Rule systematically. For our exercise, this method indeed leads to the correct expression for the derivative by simply following structured steps.
Differentiation Techniques
Differentiation Techniques are various strategies used to find derivatives, especially those expanding beyond basic rules. While we deal with functions that are products or combinations of polynomials and other forms, these special methods like the Product Rule become indispensable.
With basic differentiation, each term can be addressed independently when functions are simplified. Yet, for products, quotients, or compositions, tailored approaches like the Product Rule, Chain Rule, and Quotient Rule enter play.
  • It's about knowing the right technique for the situation.
  • Combining these methods might help simplify and efficiently solve complex derivatives.
  • Strategically choosing a technique can lead to an easier, more understandable process.
The step-by-step approach seen in our exercise highlights applying these techniques to tackle more complicated derivatives efficiently. Understand how derivatives interact under multiplication or within more complex algebraic structures, and practice is crucial. Ultimately, selecting the right differentiation technique shapes your ability to manage even the most tangled of derivatives with ease.
Polynomial Differentiation
Polynomial Differentiation deals specifically with finding the derivative of polynomial expressions. A polynomial is built by terms consisting of a coefficient and a variable raised to a power. Finding derivatives of such terms follows a basic rule: the power rule.
  • The Power Rule helps us handle terms like \( x^n \).
  • In its simplest form, it states: \( \frac{d}{dx}(x^n) = nx^{n-1} \).
So, if you have \( x^3 \), its derivative is \( 3x^2 \). This rule is applied term-by-term, even in larger expressions.
For instance, in our solved problem, terms like \( x^{3/4} \) have derivatives such as \( \frac{3}{4}x^{-1/4} \). Likewise, each polynomial component gets processed separately. Then, combine them based on simple algebra rules.
Dealing with polynomials is often straightforward due to these rules. Each term can be quickly transformed, and these transformations apply universally to each component of the polynomial. By practicing polynomial differentiation, you build a strong mathematical foundation for analyzing real-world functions and system behaviors.

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