Question

$$\text { If } s=t^{u}, \text { find } \partial s / \partial t, \partial s / \partial u$$.

Short answer

\(\partial s / \partial t = u t^{u-1}\) and \(\partial s / \partial u = t^u \ln(t)\).

Step-by-step solution

Identify the given function

A function is given in the form of \[s = t^u\] It implies that s is a function of both t and u.

Apply partial differentiation with respect to t

To find \(\partial s / \partial t\), treat u as a constant. Using the power rule for differentiation: \[\frac{\partial s}{\partial t} = \frac{\partial}{\partial t} (t^u) = u t^{u-1}\] Thus, \(\partial s / \partial t = u t^{u-1}\).

Apply partial differentiation with respect to u

To find \(\partial s / \partial u\), treat t as a constant. Recall that \(t^u = e^{u \ln(t)}\): Differentiating with respect to u, we get: \[\frac{\partial s}{\partial u} = \frac{\partial}{\partial u} (e^{u \ln(t)}) = e^{u \ln(t)} \frac{d}{du} (u \ln(t))= t^u \ln(t)\] Thus, \(\partial s / \partial u = t^u \ln(t)\).

Key concepts

Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions with more than one variable. In this exercise, we must differentiate a function of two variables, specifically, \(s = t^u\). The goal is to find the partial derivatives \(\frac{\partial s}{\partial t}\) and \(\frac{\partial s}{\partial u}\). Partial derivatives measure the rate of change of the function with respect to one variable while keeping the other variables constant.

Understanding multivariable calculus is crucial in physics, engineering, and economics, where functions often depend on several variables. For instance, the temperature \(T \) at a point in space may depend on its coordinates \((x, y, z)\), requiring the use of partial derivatives to understand how \(T \) changes with respect to each coordinate.

Let's break down the partial differentiation process step by step for our specific function.

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions. If a variable depends on another variable, which in turn depends on a third variable, the chain rule allows us to find the derivative of the first variable with respect to the third. Mathematically, if \(z = f(y)\) and \(y = g(x)\), then the chain rule states that: \[\frac{dz}{dx} = \frac{df}{dy} \cdot \frac{dy}{dx} \] In this exercise, implicitly applying the chain rule simplifies calculating partial derivatives.

For \(\frac{\partial s}{\partial t}\), we treat \(u\) as constant and use the power rule, leading to \(u t^{u-1}\). For \(\frac{\partial s}{\partial u}\), we use the fact that \(t^u = e^{u \ln(t)}\), which is easier to differentiate with respect to \(u\). This exemplifies how the chain rule aids in breaking down complicated expressions by leveraging known derivatives of simpler functions.

Implicit Differentiation

Implicit differentiation is a technique for finding derivatives of functions that are not explicitly solved for one variable in terms of others. Unlike explicit functions where \(y\) is directly given as a function of \(x\), implicit differentiation allows us to handle situations where solving the function explicitly is difficult or impossible.

Consider our function \(s = t^u\). When finding \(\frac{\partial s}{\partial u}\), we implicitly express \(t^u\) as \(e^{u \ln(t)}\), since the exponential function is straightforward to differentiate. This method involves differentiating each side of the equation with respect to the desired variable and solving for the derivative.

By using implicit differentiation, we transform \(t^u\) into a function where the exponent \(u\) is easier to handle. This approach is incredibly useful in multivariable calculus when direct differentiation seems complex or unattainable.