Problem 19 - Essential Calculus Early Transcendentals

Question

Differentiate the function. $$ f(x)=\left(x^{3}+2 x\right) e^{x} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Function Type

The function given is \( f(x) = (x^3 + 2x)e^x \). This is a product of two functions, namely \( u(x) = x^3 + 2x \) and \( v(x) = e^x \). We will use the product rule to differentiate this function.

Step 2: Differentiate the First Function

Differentiate \( u(x) = x^3 + 2x \) with respect to \( x \). This gives us \( u'(x) = 3x^2 + 2 \).

Step 3: Differentiate the Second Function

Differentiate \( v(x) = e^x \) with respect to \( x \). Since the derivative of \( e^x \) is \( e^x \), we have \( v'(x) = e^x \).

Step 4: Apply the Product Rule

The product rule states that \((uv)' = u'v + uv'\). Substituting the derivatives we found into this formula gives:\[ f'(x) = (3x^2 + 2)e^x + (x^3 + 2x)e^x \].

Step 5: Simplify the Expression

Combine and simplify the expression:\[ f'(x) = (3x^2 + 2)e^x + x^3e^x + 2xe^x \]Combine like terms to get: \[ f'(x) = (x^3 + 3x^2 + 2x + 2)e^x \].

Key Concepts

Product Rule

When differentiating the product of two functions, the Product Rule comes into play. This rule is essential when dealing with functions like our given function \( f(x) = (x^3 + 2x)e^x \), which is a multiplication of two separate simpler functions.
The Product Rule can be stated as follows:

• If you have two differentiable functions \( u(x) \) and \( v(x) \), their derivative \( (uv)' \) is given by \( u'v + uv' \).
• This means you first differentiate \( u(x) \) while keeping \( v(x) \) the same, and then differentiate \( v(x) \) while keeping \( u(x) \) the same, and finally, sum up these two products.

For the given function, we identify \( u(x) = x^3 + 2x \) and \( v(x) = e^x \). Next, we differentiate each one individually and plug them into the product rule formula to find \( f'(x) \).
By using the Product Rule, it becomes easier to tackle more complex functions where multiplication is involved, simplifying differentiation significantly.

Function Differentiation

Differentiation is the cornerstone of calculus, allowing us to find the rate at which a function changes at any given point. In our problem, we need to differentiate two functions: a polynomial function and an exponential function.
To differentiate the polynomial part \( u(x) = x^3 + 2x \):

• The derivative of \( x^3 \) with respect to \( x \) is \( 3x^2 \).
• The derivative of \( 2x \) with respect to \( x \) is \( 2 \).

Putting these results together gives us \( u'(x) = 3x^2 + 2 \).
Differentiation rules such as the power rule make simple polynomial differentiation straightforward and systematic, enabling us to approach more complex components with confidence.

Exponential Function Differentiation

Exponential functions have unique derivatives that make them powerful tools in calculus. For the function \( v(x) = e^x \), differentiation is direct yet special because the derivative of \( e^x \) is simply \( e^x \), remaining unchanged.
This characteristic simplifies differentiation of exponential functions, as you don't need to go through additional rules. For the given function, this means:

• The differentiation of \( e^x \) with respect to \( x \) is \( v'(x) = e^x \). This constant rule holds true for any coefficient of \( x \) in the exponent.

Exponential functions maintain their form after differentiation, which helps when applying the product rule, as seen here when substituting \( v'(x) = e^x \) back into the derivative expression.
Understanding exponential differentiation fully equips you to handle complex equations involving exponential growth or decay, commonly found in real-world applications.