Problem 65 - Calculus Single Variable

Question

Dawson's integral is the function defined for \(x \geq 0\) by $$ F(x)=\exp \left(-x^{2}\right) \int_{0}^{x} \exp \left(t^{2}\right) d t $$ Compute \(F^{\prime}(x)\)

Step-by-Step Solution

Answer

\( F'(x) = 1 - 2x e^{-x^2} \int_0^x e^{t^2} \ dt \)

Step 1: Understand the Function

We have the function \( F(x) = e^{-x^2} \int_0^x e^{t^2} \ dt \). We need to find the derivative \( F'(x) \).

Step 2: Apply the Product Rule

The function \( F(x) \) is a product of two functions: \( e^{-x^2} \) and \( \int_0^x e^{t^2} \ dt \). Therefore, we'll use the product rule, which states if \( y = u(x)v(x) \), then \( y' = u'v + uv' \).

Step 3: Differentiate the First Function

Differentiate \( e^{-x^2} \) with respect to \( x \). Use the chain rule: if \( u = e^{f(x)} \), then \( u' = e^{f(x)}f'(x) \).\[ \frac{d}{dx}(e^{-x^2}) = e^{-x^2}(-2x) = -2xe^{-x^2} \]

Step 4: Differentiate the Integral using Leibniz's Rule

The Leibniz integral rule helps differentiate an integral where the upper limit is a variable:\[ \frac{d}{dx} \left( \int_0^x e^{t^2} \, dt \right) = e^{x^2}\]

Step 5: Utilize the Product Rule

Using the product rule:\[ F'(x) = \left(e^{-x^2} \int_0^x e^{t^2} \, dt \right)' = \left(e^{-x^2}\right)' \cdot \int_0^x e^{t^2} \, dt + e^{-x^2} \cdot \frac{d}{dx} \left( \int_0^x e^{t^2} \, dt \right) \]Substitute the derivatives from Step 3 and 4:\[ F'(x) = (-2xe^{-x^2}) \cdot \int_0^x e^{t^2} \, dt + e^{-x^2} \cdot e^{x^2} \]

Step 6: Simplify the Expression

The expression simplifies as follows:\[ F'(x) = -2x e^{-x^2} \int_0^x e^{t^2} \, dt + 1 \]The terms \( e^{-x^2} \cdot e^{x^2} \) cancel to 1.

Key Concepts

Product Rule

The product rule is an essential tool in calculus for finding the derivative of a product of two functions. When faced with a function that is the product of two separate functions, such as \( F(x) = u(x) \cdot v(x) \), the product rule simplifies the process of differentiation.
When applying the product rule, we use the formula:

• If \( y = u(x) \cdot v(x) \), then the derivative \( y' = u'(x) \cdot v(x) + u(x) \cdot v'(x) \).

In our case with Dawson's integral:

• The functions are \( e^{-x^2} \) and \( \int_0^x e^{t^2} \ dt \).
• We differentiate each of these functions separately and apply the product rule to combine them.

This method ensures that we correctly compute the changes in both parts of the product, crucial for calculating accurate derivatives of complex functions.

Chain Rule

The chain rule is a powerful technique in calculus used to find the derivative of a composite function. Whenever you have a function within another function, the chain rule is typically the way to go.
Let's say you have a function \( u = f(g(x)) \), the derivative is then given by:

• \( u' = f'(g(x)) \cdot g'(x) \).

In the problem of Dawson's integral, we have the function \( e^{-x^2} \). To find its derivative, we use the chain rule:

• Inner function: \( g(x) = -x^2 \).
• Outer function: \( f(u) = e^u \).

Thus, differentiating gives us:

• \( \frac{d}{dx}(e^{-x^2}) = e^{-x^2}(-2x) = -2x e^{-x^2} \).

This calculation is pivotal when multiplying it by another function as seen in the product rule above.

Leibniz Integral Rule

The Leibniz integral rule is essential when differentiating certain types of integrals that have variable limits. This rule is particularly useful when the differentiation has to be performed under the integral sign, as is needed in Dawson's integral problem.
The Leibniz rule can be stated as:

• \( \frac{d}{dx} \left( \int_{a(x)}^{b(x)} f(t, x) \, dt \right) = f(b(x), x) \cdot b'(x) - f(a(x), x) \cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x}f(t, x) \ dt \).

For Dawson’s integral:

• The integral is \( \int_0^x e^{t^2} \, dt \).
• Using the Leibniz rule, it simplifies to: \( \frac{d}{dx} \left( \int_0^x e^{t^2} \, dt \right) = e^{x^2} \), as the lower bound is constant and thus its derivative vanishes.

This result helps us find how fast the integral changes with respect to \( x \), key to resolving our problem.

Calculus Differentiation

Calculus differentiation is the process by which we determine the rate at which a function is changing at any point. It's a foundational concept in calculus and is crucial for understanding the behavior of functions.
Differentiation allows us to:

• Compute tangents to curves, indicating slope or rate of change.
• Understand and solve complex systems, often modelled by mathematical functions.

In the context of Dawson's integral, differentiation is employed through product rule, chain rule, and the application of the Leibniz integral rule. This rich combination provides a holistic view of how the function behaves as it incorporates the change dynamic of each element in the expression.
Mastering differentiation techniques helps tackle intricate problems like computing \( F'(x) \) efficiently, offering insights into how composite functions grow or decrease.