Problem 18 - Single Variable Calculus

Question

Find the differential of the function at the indicated number. $$ f(x)=2^{-x^{2}} ; \quad x=1 $$

Step-by-Step Solution

Answer

The differential of the function \(f(x)=2^{-x^2}\) at \(x=1\) is \(\frac{df}{dx}(1) = -\ln(2)\).

Step 1: Find the derivative of the function

To find the derivative of the function \(f(x) = 2^{-x^2}\), we can use the chain rule. The chain rule states that \[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \] In this case, let \(f(u) = 2^u\) and \(g(x) = -x^2\). Then we have \(f(g(x)) = 2^{-x^2}\). Now, we will find the derivative of \(f(u)\) and \(g(x)\). \[ f'(u) = \frac{d}{du}(2^u) = 2^u \ln(2) \] \[ g'(x) = \frac{d}{dx}(-x^2) = -2x \] Using the chain rule, we can now find the derivative of our function. \[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) = 2^{-x^2} \ln(2) \cdot (-2x) \] Therefore, the derivative of the function is \(\frac{df}{dx} = 2^{-x^2} \ln(2)(-2x)\).

Step 2: Evaluate the derivative at the given point

Now, we need to find the value of the derivative at \(x = 1\). We substitute \(x = 1\) into the derivative: \[ \frac{df}{dx}(1) = 2^{-1^2} \ln(2)(-2(1)) = 2^{-1} \ln(2)(-2) = \frac{1}{2} \ln(2)(-2) \] So the derivative of the function at the point \(x = 1\) is \(\frac{df}{dx}(1) = -\ln(2)\).

Key Concepts

Chain Rule

Mastering the chain rule is essential for any student diving into differential calculus. Think of it as a way to peel back the layers of an onion. When you have a function that's nested within another function, like a Russian doll, the chain rule helps you differentiate these composite functions. It essentially states that if you have a function inside another function—let's say, f inside g—you differentiate the outer function while keeping the inner one as is, and then multiply by the derivative of the inner function.

In our example, we had the function f(x) = 2^{-x^2}. Imagining that as an onion, the outermost layer is 2^u, and inside it, taking the place of u, is our second layer, -x^2. Using the chain rule, we differentiate the outer layer with respect to u, then we multiply that by the derivative of our inner layer, -x^2, with respect to x.

Derivative of Exponential Functions

Now, diving into exponential functions, these are the ones that look like a^x, where a is a constant and x is the exponent. When we differentiate these functions, a new component comes into play—the natural logarithm, denoted as ln.

The derivative of an exponential function a^x is a^x ln(a). Why is that? It's because the natural logarithm is the inverse of the exponential function with base e, the Euler's number. And in differentiation, the natural logarithm helps us adjust the rate of change when the base is not e. In the case of our function 2^u where u=-x^2, after applying this rule, we find that its derivative is 2^u ln(2), reflecting the relationship between exponentials and logarithms.

Evaluating Derivatives

Once we've calculated the derivative—or the rate at which our function changes—we might need to evaluate it at a particular point to understand the behavior of the function at that specific location. Evaluating derivatives is basically plugging in a value for x and calculating the result.

To evaluate the derivative at a given point, we just replace x with the number we're interested in. If our derived formula is f'(x) = 2^{-x^2}ln(2)(-2x) and we need to know the value at x=1, we substitute 1 for every x in the derivative. This allows us to determine the instantaneous rate of change of our function at x=1, effectively giving us a snapshot of the function's behavior at that very point. For deeper understanding, students should practice substituting different values into the derivative, which allows them to explore how the function changes at various points along its domain.