Problem 51 - Calculus Early Transcendentals

Question

Compute the following derivatives using the method of your choice. \(\frac{d}{d x}\left(e^{-10 x^{2}}\right)\)

Step-by-Step Solution

Answer

Answer: The derivative of the function is \(\frac{d}{dx}\left(e^{-10x^2}\right) = -20xe^{-10x^2}\).

Step 1: Find f'(u) and g'(x)

Let's first find the derivative of f(u) with respect to u. Since f(u) = e^u, we have f'(u) = e^u. Now let's find the derivative of g(x) with respect to x. Since g(x) = -10x^2, we have g'(x) = -20x.

Step 2: Apply the chain rule

Now that we have f'(u) and g'(x), we can apply the chain rule to find the derivative of y(x). According to the chain rule, we have \[y'(x) = f'(g(x)) * g'(x).\] Substitute the expressions for f'(u) and g'(x) into the equation and replace u with g(x): \[y'(x) = e^{g(x)} * (-20x).\] Notice that g(x) = -10x^2, so we can substitute this into the above expression: \[y'(x) = e^{-10x^2} * (-20x).\] We have found the derivative of the function y(x): \[\frac{d}{dx}\left(e^{-10x^2}\right) = -20xe^{-10x^2}.\]

Key Concepts

Chain Rule

In calculus, the chain rule is a powerful tool for finding the derivative of composite functions. Imagine it like peeling an onion: you address each layer individually before capturing the entire essence. For a function composed of two functions, say \(y(x) = f(g(x))\), the chain rule states that:

• Find the derivative of the outer function \(f\) with respect to \(g(x)\) (not \(x\)), which is \( f'(g(x)) \).
• Next, find the derivative of the inner function \(g(x)\) with respect to \(x\), which is \( g'(x) \).
• Finally, multiply these two derivatives together: \( y'(x) = f'(g(x)) \cdot g'(x) \).

Using the original problem, we applied the chain rule by treating \( e^{-10x^2} \) as a composition of \(f(u) = e^u\) and \(g(x) = -10x^2\). Thus, the derivative was calculated as:\[ y'(x) = e^{-10x^2} \cdot (-20x) = -20x e^{-10x^2}. \]

Derivative of Exponential Functions

Exponential functions have the form \(e^x\), where \(e\) is a constant approximately equal to 2.71828. The unique property of these functions is that their derivatives are proportional to the original function: \( \frac{d}{dx} e^x = e^x \).In more complex expressions involving exponentials, like \(e^{u(x)}\), you leverage the same rule, but you also account for the inner function using the chain rule. Thus, the derivative becomes:

• Apply the chain rule: \( \frac{d}{dx} e^{u(x)} = e^{u(x)} \cdot u'(x) \).

For our original exercise, the inner function was \(-10x^2\), whose derivative is \(-20x\). Thus, the derivative of \(e^{-10x^2}\) used the form:\[ \frac{d}{dx} e^{-10x^2} = e^{-10x^2} \cdot (-20x) = -20x e^{-10x^2}. \]

Product Rule

The product rule is used to differentiate expressions where two or more functions are multiplied together. It states that for two functions \(u(x)\) and \(v(x)\), the derivative of their product is given by:

• \( (uv)' = u'v + uv' \).

Though not directly used in this particular problem, understanding the product rule is crucial when dealing with functions that involve products of expressions. For example, if we had tackled a problem involving multiplying two different functions together, we could apply this rule effectively.In our exercise, it simply reinforced the structured approach to breaking down derivatives into manageable parts, especially when splitting an exponential and polynomial term was necessary for conceptual clarity. This approach helps in maintaining accuracy while computing derivatives.