Problem 37 - Calculus - AP Edition

Question

Find \(d y / d x\) using any method. $$y=\left(x^{3}-2 x^{2}+1\right) e^{x}$$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = e^x(x^3 + x^2 - 4x + 1)\).

Step 1: Identify the Rule

We need to find the derivative of the function \(y=(x^3-2x^2+1)e^x\). This is a product of two functions, so we will use the product rule to find \(\frac{dy}{dx}\). The product rule states that if \(u(x)\) and \(v(x)\) are functions, \((uv)' = u'v + uv'\).

Step 2: Define the Functions

Identify the functions as \(u(x) = x^3 - 2x^2 + 1\) and \(v(x) = e^x\). This means \(y = u(x)v(x)\).

Step 3: Differentiate \(u(x)\)

Find the derivative of \(u(x) = x^3 - 2x^2 + 1\). Using the power rule, \(u'(x) = 3x^2 - 4x\).

Step 4: Differentiate \(v(x)\)

Since \(v(x) = e^x\), its derivative is \(v'(x) = e^x\).

Step 5: Apply the Product Rule

Substitute \(u(x)\), \(u'(x)\), \(v(x)\), and \(v'(x)\) into the product rule formula: \[ \frac{dy}{dx} = (3x^2 - 4x)e^x + (x^3 - 2x^2 + 1)e^x. \]

Step 6: Simplify

Factor out \(e^x\) from both terms: \[ \frac{dy}{dx} = e^x(3x^2 - 4x + x^3 - 2x^2 + 1). \] Simplify the expression inside the parenthesis: \(x^3 + (3x^2 - 2x^2) - 4x + 1 = x^3 + x^2 - 4x + 1\). Thus, \[ \frac{dy}{dx} = e^x(x^3 + x^2 - 4x + 1). \]

Key Concepts

Product Rule

The product rule is a fundamental technique in calculus for finding the derivative of a product of two functions. Imagine you have two functions, named \(u(x)\) and \(v(x)\). When you multiply these functions, the result is another function, say \(y(x) = u(x)v(x)\). To find the derivative of \(y(x)\), you cannot simply differentiate each function separately and multiply the results. Instead, you must use the product rule formula, which states that the derivative \((uv)'\) is:

• \(u'(x)v(x) + u(x)v'(x)\)

To apply this rule:

• First, differentiate \(u(x)\) to get \(u'(x)\).
• Next, differentiate \(v(x)\) to get \(v'(x)\).
• Multiply \(u'(x)\) by \(v(x)\) and add it to \(u(x)\) multiplied by \(v'(x)\).

This rule is particularly helpful when dealing with problems like \(y=(x^3-2x^2+1)e^x\), where both parts of the multiplication—\(x^3-2x^2+1\) and \(e^x\)—need to be treated as separate functions before combining their derivatives.

Power Rule

In calculus, the power rule is a straightforward formula used to differentiate functions of the form \(x^n\). Here, \(n\) is any real number. The power rule tells us that the derivative of \(x^n\) is \(nx^{n-1}\). It's like lowering the power by one and multiplying by the original exponent.For example:

• To find the derivative of \(x^3\), apply the power rule: \(3x^{3-1} = 3x^2\).
• Similarly, for \(-2x^2\), using the power rule gives \(-2(2)x^{2-1} = -4x\).

In the context of the exercise, to find the derivative of \(u(x) = x^3 - 2x^2 + 1\), we use the power rule:

• The derivative of \(x^3\) is \(3x^2\).
• The derivative of \(-2x^2\) is \(-4x\).
• The derivative of a constant like \(1\) is \(0\), due to constants having no rate of change.

Thus, the derivative of \(u(x)\) becomes \(3x^2 - 4x\). The power rule simplifies the differentiation process, enabling you to handle algebraic functions efficiently.

Exponential Function

Exponential functions, especially those involving \(e\), are incredibly important in calculus and many other fields due to their unique properties. The function \(e^x\), where \(e\) is the base of the natural logarithm, approximately equal to 2.718, is a classic example of an exponential function. One of the key features of the function \(e^x\) is that its derivative is also \(e^x\). This simplicity makes it quite unique among exponential functions. When differentiating \(e^x\), you find:

• The derivative \(\frac{d}{dx}[e^x] = e^x\)

In the exercise, \(v(x) = e^x\), so its derivative, \(v'(x)\), is naturally \(e^x\) as well. This property is handy when using the product rule because it means you can maintain this part of the function unchanged during differentiation. Thus, when you combine it with other derivatives, it does not complicate the resulting function further, keeping the differentiation process straightforward.