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Chapter 3: Problem 5

Differentiate the functions. $$y=x\left(x^{2}+1\right)^{4}$$

Short Answer

Expert verified
The derivative is \( (x^2 + 1)^4 + 8x^2 (x^2 + 1)^3 \).

Step by step solution

01

Identify the Parts of the Function

The function is given by \( y = x (x^2 + 1)^4 \). This function is a product of two functions, where \( u = x \) and \( v = (x^2 + 1)^4 \). We will use the product rule to differentiate the function.
02

State the Product Rule

The product rule states that \( (uv)' = u'v + uv' \). We need to find \( u' \) and \( v' \).
03

Differentiate Each Part

First, find \( u' \). Since \( u = x \), \( u' \) is the derivative of \( x \), which is \( 1 \). Next, we find \( v' \). To differentiate \( v = (x^2 + 1)^4 \), use the chain rule.
04

Apply the Chain Rule

To differentiate \( v = (x^2 + 1)^4 \), let \( w = x^2 + 1 \). Then, \( v = w^4 \). The derivative \( v' \) is \( 4w^3 \cdot w' \). Since \( w = x^2 + 1 \), \( w' = 2x \). Thus, \( v' = 4(x^2 + 1)^3 \cdot 2x \). This simplifies to \( v' = 8x (x^2 + 1)^3 \).
05

Combine Using the Product Rule

Using the product rule, \( y' = u'v + uv' \). Substituting the values we have found, \( y' = 1 \times (x^2 + 1)^4 + x \times 8x (x^2 + 1)^3 \).
06

Simplify the Expression

Simplify the expression: \( y' = (x^2 + 1)^4 + 8x^2 (x^2 + 1)^3 \).

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Key Concepts

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

Product Rule
The \textbf{Product Rule} is a fundamental concept in differentiation, helping determine the derivative of a product of two functions. It works because the change in a product function depends on the changes in both of its component functions. The rule states that for two functions, \(u\) and \(v\), the derivative of their product is given by: \[ (uv)' = u'v + uv' \].

Here, we identified \(u = x\) and \(v = (x^2 + 1)^4\).

First, find the derivative of \(u\), which is straightforward:
  • \( u = x \)
  • \( u' = 1 \)
Next, we moved on to finding the derivative of \(v\). To do this, we needed the chain rule.
Chain Rule
The \textbf{Chain Rule} is another critical tool for finding the derivative of composite functions. When you have a function inside another function, the chain rule helps to differentiate it systematically. It can be summarized as: \[ (f(g(x)))' = f'(g(x)) \times g'(x) \].

For our exercise, we needed to differentiate \(v = (x^2 + 1)^4\). To apply the chain rule, we let:
  • \( w = x^2 + 1 \)
  • \( v = w^4 \)
Now, differentiate \(w\) and \(v\):
  • \( w' = 2x \)
  • \( v' = 4w^3 \times w' \)
Substituting \(x^2 + 1\) back for \(w\):
  • \( v' = 4(x^2 + 1)^3 \times 2x \)
  • \( v' = 8x (x^2 + 1)^3 \)
With \(u'\) and \(v'\) known, we can combine them using the product rule.
Simplifying Derivatives
The final step in differentiation often involves \textbf{Simplifying Derivatives} for clarity and utility. Simplification makes the result easier to interpret and use in subsequent calculations. In our case, after applying the product rule:\[ y' = u'v + uv' \, \text{where} \, u' = 1 \, \text{and} \, v' = 8x (x^2 + 1)^3 \] Combining what we have: \[ y' = 1 \times (x^2 + 1)^4 + x \times 8x (x^2 + 1)^3 \] Then, we simplify this expression further: \[ y' = (x^2 + 1)^4 + 8x^2 (x^2 + 1)^3 \] Now, you have a clear, simplified derivative, which is easier to analyze and use.

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Most popular questions from this chapter

Suppose that \(x\) and \(y\) are related by the given equation and use implicit differentiation to determine \(\frac{d y}{d x}.\) $$x^{2}+4 x y+4 y=1$$

Find the point(s) on the graph of \(y=\left(x^{2}+3 x-1\right) / x\) where the slope is \(5.\)

Suppose that \(x\) and \(y\) are both differentiable functions of \(t\) and are related by the given equation. Use implicit differentiation with respect to \(t\) to determine \(\frac{d y}{d t}\) in terms of \(x, y\) and \(\frac{d x}{d t}\). $$x^{2}+2 x y=y^{3}$$

A function \(h(x)\) is defined in terms of a differentiable \(f(x) .\) Find an expression for \(h^{\prime}(x).\) $$h(x)=-f(-x)$$

A function \(h(x)\) is defined in terms of a differentiable \(f(x) .\) Find an expression for \(h^{\prime}(x).\) $$h(x)=f(f(x))$$
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