Problem 23 - Thomas Calculus

Question

In Exercises \(17-36,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\left(x^{2}-2 x+2\right) e^{x} $$

Step-by-Step Solution

Answer

The derivative is \( y' = x^2 e^x \).

Step 1: Identify the Functions

The given function is a product of two functions: \( f(x) = x^2 - 2x + 2 \) and \( g(x) = e^x \). We need to find the derivative using the product rule.

Step 2: Recall the Product Rule Formula

The derivative of a product of two functions \( u(x) \) and \( v(x) \) is given by \( (uv)' = u'v + uv' \). We'll apply this rule to our functions \( f(x) \) and \( g(x) \).

Step 3: Find the Derivative of the First Function

Compute \( f'(x) \) where \( f(x) = x^2 - 2x + 2 \). By differentiating, we get: \( f'(x) = 2x - 2 \).

Step 4: Find the Derivative of the Exponential Function

Compute \( g'(x) \) where \( g(x) = e^x \). The derivative of \( e^x \) is itself: \( g'(x) = e^x \).

Step 5: Apply the Product Rule

Using the product rule: \( (f(x)g(x))' = f'(x)g(x) + f(x)g'(x) \). Substitute the derivatives we found: \[ (x^2 - 2x + 2)' e^x + (x^2 - 2x + 2)(e^x)' \] simplifies to \[ (2x - 2)e^x + (x^2 - 2x + 2)e^x \].

Step 6: Simplify the Expression

Combine the terms: \( (2x - 2 + x^2 - 2x + 2)e^x = x^2 e^x \). This is the simplified derivative with respect to \( x \).

Key Concepts

Product Rule

In calculus, the product rule is an essential technique used to find the derivative of a product of two functions. This rule is applied when you have two differentiable functions multiplied together, and you want to find their combined derivative. For functions \( u(x) \) and \( v(x) \), the product rule is stated as:

• \((uv)' = u'v + uv'\)

This means that you differentiate the first function and multiply it by the second function, then add the product of the first function with the derivative of the second function.
Let's see how this applies in practice:

• Function \( f(x) = x^2 - 2x + 2 \) and \( g(x) = e^x \).
• The derivative of their product is \( (f(x)g(x))' \).
• Applying the product rule, it becomes \( f'(x)g(x) + f(x)g'(x) \).

This helps you systematically break down the problem into manageable parts.

Exponential Function Differentiation

Differentiating exponential functions, especially those involving \( e \), is a common task in calculus. The beauty of the exponential function, \( e^x \), is that its derivative is the same as the function itself. That is:

• \( \frac{d}{dx}e^x = e^x \)

This property makes exponential functions straightforward to work with during differentiation.
Let's consider the function \( g(x) = e^x \). Differentiating it results in \( g'(x) = e^x \), which matches the original function.
In our example, when using the product rule:

• The derivative of the exponential part, \( e^x \), doesn't change, simplifying the calculation.
• This helps maintain clarity when multiplying with the other function's derivatives.

Remember, this property makes the calculations more direct and error-free.

Function Derivative

Taking the derivative of individual functions is the preliminary step in applying calculus rules like the product rule. For the polynomial function \( f(x) = x^2 - 2x + 2 \), you differentiate using basic power rules:

• For \( x^n \), the derivative is \( nx^{n-1} \).

Applying this:

• The derivative of \( x^2 \) is \( 2x \).
• The derivative of \( -2x \) is \( -2 \).
• The derivative of a constant, \( 2 \), is \( 0 \).

So, \( f'(x) = 2x - 2 \).
This derivative is used in conjunction with \( g'(x) = e^x \) when applying the product rule.
Understanding how to find function derivatives allows you to tackle more complex problems systematically. Take care with each step, and ensure calculations are precise for accurate results.