Question

Find \(D_{x} y\) using the rules of this section. $$ y=x\left(x^{2}+1\right) $$

Short answer

The derivative is \(3x^2 + 1\).

Step-by-step solution

Identify the Rule to Apply

In this problem, we have to differentiate a product of two functions, specifically, \(y = x(x^2 + 1)\). The rule we'll use is the product rule, which states that if \(y = uv\), then \(D_x y = u'v + uv'\), where \(u = x\) and \(v = x^2 + 1\).

Differentiate Each Function

To apply the product rule, we need the derivatives \(u'\) and \(v'\). Start with \(u = x\), so \(u' = \frac{d}{dx} (x) = 1\). Next, take the derivative of \(v = x^2 + 1\), which gives \(v' = \frac{d}{dx} (x^2 + 1) = 2x\).

Apply the Product Rule

Now substitute \(u\), \(v\), \(u'\) and \(v'\) into the product rule formula: \(D_x y = u'v + uv' = 1(x^2 + 1) + x(2x)\).

Simplify the Expression

Expand the terms from the product rule: \(1(x^2 + 1) = x^2 + 1\) and \(x(2x) = 2x^2\). Now, combine these results: \(D_x y = x^2 + 1 + 2x^2\).

Combine Like Terms

Combine the like terms in the expression to simplify: \(D_x y = 3x^2 + 1\).

Key concepts

product rule

The product rule is an essential technique in calculus differentiation. It is used when you need to find the derivative of a function that is the product of two simpler functions. When you have a composite function like \( y = uv \), where \( u\) and \( v \) are both functions of \( x \), you apply the product rule.

Here's how it works: the derivative of the product \( uv \) is found by differentiating \( u \) while keeping \( v \) constant, and then adding this to the product of \( u \) and the derivative of \( v \). Mathematically, it is expressed as:

• \( D_x y = u'v + uv' \)

This rule is crucial because it can be a bit tricky to manage the derivation of products directly. Remember, it's all about breaking down the functions into manageable parts and processing each derivative part separately, then reassembling them neatly. Once you grasp the underlying logic, the product rule becomes a powerful tool in your calculus toolkit.

derivative examples

Understanding differentiation through examples is vital because it allows you to see how rules like the product rule are applied in practice.

Consider the function \( y = x(x^2 + 1) \). Here, it represents the product of two simpler functions: \( u = x \) and \( v = x^2 + 1 \).

• First, find the derivative of \( u \), which is \( u' = \frac{d}{dx}(x) = 1 \).
• Then, find the derivative of \( v \), which is \( v' = \frac{d}{dx}(x^2 + 1) = 2x \).

Applying the product rule to these functions, substituting into \( D_x y = u'v + uv' \), yields:

• \( D_x y = 1(x^2 + 1) + x(2x) \)

By expanding and adding these expressions, you simplify to \( D_x y = 3x^2 + 1 \). This example shows how separate derivatives come together to form the derivative of the combined function, highlighting the step-by-step application of differentiation techniques.

differentiation techniques

Differentiation techniques form the backbone of calculus and allow us to analyze how functions change. They help calculate the slope of a function at any point and provide insight into the function's behavior.

There are several elementary rules, like the power rule, sum rule, and most importantly for us, the product rule, each of which finds use under different scenarios:

• **Power Rule:** Differentiate functions of the form \( x^n \).
• **Sum/Difference Rule:** Handle sums or differences of functions.
• **Product Rule:** Manage multiplication of two functions.

Each rule has its place, and frequently we use a combination of these rules. For example, the differentiation of a polynomial might require the power and sum rules.

The ability to choose the right rule or combination of rules is key to efficient and correct differentiation. Mastering these becomes easier with practice, building up intuition on how to handle diverse types of functions correctly. Remember, each step in differentiation is about breaking down complex expressions into simpler parts and applying the specific rules methodically.