Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-16, find all first partial derivatives of each function. \(f(s, t)=e^{t^{2}-s^{2}}\)

Step-by-Step Solution

Answer

The partial derivatives are \(f_s = -2s \, e^{t^2-s^2}\) and \(f_t = 2t \, e^{t^2-s^2}\).

Step 1: Identify Variables

The function given is \(f(s, t)=e^{t^{2}-s^{2}}\). We need to find the partial derivative with respect to each variable: \(s\) and \(t\).

Step 2: Partial Derivative with respect to \(s\)

To find \(f_s\), the partial derivative of \(f\) with respect to \(s\), we treat \(t\) as a constant and differentiate \(e^{t^{2}-s^{2}}\) with respect to \(s\). Using the chain rule, the derivative of \(e^{u}\) is \(e^{u} \cdot \frac{du}{ds}\), where \(u = t^2 - s^2\). Thus, \(\frac{du}{ds} = -2s\). Therefore, \(f_s(s, t)=e^{t^{2}-s^{2}} \, (-2s) = -2s \, e^{t^{2}-s^{2}}\).

Step 3: Partial Derivative with respect to \(t\)

To find \(f_t\), the partial derivative of \(f\) with respect to \(t\), we treat \(s\) as a constant and differentiate \(e^{t^{2}-s^{2}}\) with respect to \(t\). Again, using the chain rule, \(u = t^2 - s^2\), hence \(\frac{du}{dt} = 2t\). Therefore, \(f_t(s, t)=e^{t^{2}-s^{2}} \, (2t) = 2t \, e^{t^{2}-s^{2}}\).

Key Concepts

Chain Rule

The chain rule is an essential concept in calculus, especially when dealing with functions of multiple variables. It allows us to differentiate composite functions efficiently. In simple terms, the chain rule helps find the derivative of a function that is "nested" inside another function.

Imagine you have a function that depends on another variable, like in our exercise with the function \(f(s, t) = e^{t^2 - s^2}\). When computing the derivative, we often encounter expressions where one variable affects another indirectly through a third variable. The chain rule simplifies this by relating the derivatives of these expressions.

When applying the chain rule, follow these straightforward steps:

• Find the inner function and the outer function in the composite function.
• Differentiate the outer function with respect to the inner function as if the inner function were a single variable.
• Multiply this result by the derivative of the inner function.

In the case of our example, \(u = t^2 - s^2\), which is the inner function, and \(e^u\) is the outer function. We successfully applied the chain rule for partial derivatives with respect to both \(s\) and \(t\), leading to the expressions \(-2s \, e^{t^2 - s^2}\) and \(2t \, e^{t^2 - s^2}\).

Multivariable Calculus

Multivariable calculus extends single-variable calculus concepts to functions of more than one variable. This is crucial in various scientific fields, as it allows us to analyze systems with multiple changing factors. In multivariable calculus, functions like \(f(s, t)\) in our original exercise are common.

To study such functions, we need tools like partial derivatives that isolate the change in the function with respect to each variable independently. In essence, they help us understand how the function behaves in different directions. While single-variable calculus deals with curves, multivariable calculus often involves surfaces and volumes.

Partial derivatives are calculated similarly to regular derivatives but require careful attention. We must "hold other variables constant" while differentiating regarding a specific variable. For instance, when finding \(f_s(s, t)\), \(t\) is treated as a constant, making it easier to imagine navigating a landscape where all other slopes are fixed while focus shifts to changing towards \(s\).

Differentiation

Differentiation in calculus is the process of finding a derivative, which quantifies how a function changes with respect to a variable. It reveals the exact rate of change or the slope of the tangent line at any point on the graph of the function. Differentiation is not limited to single-variable scenarios and is applied effectively in multivariable calculus to functions of more than one variable, like \(f(s, t) = e^{t^2 - s^2}\).

In the context of our exercise, differentiation helps us determine how the function \(f\) changes as each of its variables changes. This is where the concept of first partial derivatives enters. By calculating these derivatives, we obtain insight into how sensitive the function is to changes in each variable independently.

Let's recap the fundamental differentiation rules useful for this:

• The derivative of \(e^u\) with respect to \(u\) is itself, \(e^u\), multiplied by the derivative of \(u\).
• The power rule states that the derivative of \(x^n\) is \(nx^{n-1}\).
• The sum and difference rules let us differentiate terms separately and sum or subtract them.

These rules are invaluable while finding derivatives of multivariable functions as seen in our solution, leading seamlessly to the partial derivates calculated.