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In Exercises 1 through \(8,\) use the product rule to find the derivative. $$ e^{x}\left(x^{2}-3\right) $$

Short Answer

Expert verified
The derivative is \( e^x(x^2 + 2x - 3) \).

Step by step solution

01

Identify the functions

The expression given is a product of two functions. The first function is \( u(x) = e^x \) and the second function is \( v(x) = x^2 - 3 \). These are the functions we will use to apply the product rule.
02

Recall the product rule

The product rule states that the derivative of a product of two functions \( u(x) \) and \( v(x) \) is given by \( (uv)' = u'v + uv' \).
03

Differentiate each function

To use the product rule, we need the derivatives of both functions:\( u'(x) = \frac{d}{dx}(e^x) = e^x \)\( v'(x) = \frac{d}{dx}(x^2 - 3) = 2x \).
04

Apply the product rule

Substitute the functions and their derivatives into the product rule formula:\( (uv)' = e^x(x^2 - 3) + e^x(2x) \).
05

Simplify the expression

Combine and simplify the terms:\( e^x(x^2 - 3) + e^x(2x) = e^x(x^2 - 3 + 2x) \).Further simplification gives:\( = e^x(x^2 + 2x - 3) \).

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Key Concepts

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

Derivative
A derivative represents the rate of change of a function regarding one of its variables. In simple terms, it tells us how a function's value changes as its input changes. Think of it like measuring how fast a car speeds up or slows down. With derivatives, mathematicians can find the slope of a curve at any point. In the context of our problem, the derivative helps us understand the rate of change of the function \(e^{x}(x^{2}-3)\).
What we need to do first is find the derivative of each part of the function that composes the product. Then we use the product rule, a formula designed specifically to differentiate products of two or more functions.
By applying the product rule, we accurately determine how those functions interact as their variables change, providing a complete picture of the overall rate of change of the product.
Differentiation
Differentiation is the process used to find the derivative of a function. It's like finding the detailed instruction manual on how a function changes. When using differentiation, we have tools like rules and formulas that simplify the task, one of them being the product rule.
To differentiate the product example \(e^{x}(x^{2}-3)\), we broke the task into steps: recognizing the product, recalling the product rule, finding individual derivatives, and then applying the rule.
  • First, we identified the functions within the product.
  • Next, each function's derivative was determined.
  • Finally, substituting into the product rule to get the derivative of the overall product.
This ordered approach simplifies the differentiation process and ensures all parts are accounted for, providing precise results for how combined functions change.
Exponential Functions
Exponential functions are characterized by the constant base \(e\), where \(e\approx 2.718\), raised to the power of a variable. They model growth and decay processes wonderfully. Their unique property in calculus is that their derivative is the same as the function itself.
In the problem, the function \(e^x\) was one of the parts of our product. When we differentiate \(e^x\), the result is again \(e^x\). This recurrent nature simplifies many calculations.
As we applied the product rule, the exponential part ensured the process remained straightforward, adding to its value in modeling natural phenomena like population growth and radioactive decay.
Polynomial Functions
Polynomial functions, like \(x^2 - 3\) in our problem, are expressions made up of variables raised to whole-number powers, with coefficients attached. They can represent a wide variety of shapes in graphs, from simple lines to complex curves.
When differentiating polynomials, each term's derivative can be found separately. For the term \(x^2\), applying the power rule gives the derivative as \(2x\). Constant terms, like \(-3\), have a derivative of zero since they don't change.
As we applied the product rule in this exercise, the differentiation of the polynomial part provided critical changes as \(x\) varied. Understanding how these derivatives contribute when combined with those of exponential functions gives a powerful tool to analyze complex mathematical relationships.

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