Question

Find \(\partial w / \partial t\) by using the Chain Rule. Express your final answer in terms of \(s\) and \(t\) $$ w=e^{x y+z} ; x=s+t, y=s-t, z=t^{2} $$

Short answer

\( \frac{\partial w}{\partial t} = 2t(t-1)e^{s^2} \)

Step-by-step solution

Identify the Chain Rule Components

To find the partial derivative \( \frac{\partial w}{\partial t} \) using the chain rule, recognize that \( w \) is a function of \( x, y, z \), which are in turn functions of \( s \) and \( t \). Thus,\[\frac{\partial w}{\partial t} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial t} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial t}\]We need to determine these component derivatives.

Calculate \( \frac{\partial w}{\partial x} \), \( \frac{\partial w}{\partial y} \), and \( \frac{\partial w}{\partial z} \)

The function \( w = e^{xy + z} \) has partial derivatives:\[\frac{\partial w}{\partial x} = y e^{xy + z}, \quad \frac{\partial w}{\partial y} = x e^{xy + z}, \quad \frac{\partial w}{\partial z} = e^{xy + z}\]These derivatives follow from the fact that the derivative of \( e^u \) with respect to \( u \) is \( e^u \), and the use of the power rule.

Calculate the Partial Derivatives of \( x, y, z \) with respect to \( t \)

Now, find the partial derivatives:\[\frac{\partial x}{\partial t} = 1, \quad \frac{\partial y}{\partial t} = -1, \quad \frac{\partial z}{\partial t} = 2t\]These results are obtained directly from the functions: \( x = s + t \), \( y = s - t \), \( z = t^2 \).

Substitute Partial Derivatives into the Chain Rule Equation

Substitute the results from Steps 2 and 3 into the chain rule equation from Step 1:\[\frac{\partial w}{\partial t} = y \cdot e^{xy + z} \cdot 1 + x \cdot e^{xy + z} \cdot (-1) + e^{xy + z} \cdot (2t)\]Simplifying, the expression becomes:\[\frac{\partial w}{\partial t} = (y - x + 2tz)e^{xy + z}\]

Substitute Back the Expressions for \( x, y, z \) and Simplify

Replace \( x = s + t \), \( y = s - t \), \( z = t^2 \):\[\frac{\partial w}{\partial t} = ((s - t) - (s + t) + 2t^2)e^{(s+t)(s-t) + t^2}\]Simplify the terms:\[\frac{\partial w}{\partial t} = (-2t + 2t^2)e^{s^2 - t^2 + t^2} = 2t(t - 1)e^{s^2}\]

Conclusion: Final Expression for \( \frac{\partial w}{\partial t} \)

The substituting and simplifying leads us to the final expression:\[\frac{\partial w}{\partial t} = 2t(t-1)e^{s^2}\]This is the expression for the partial derivative \( \frac{\partial w}{\partial t} \) in terms of \( s \) and \( t \).

Key concepts

Partial Derivatives

Partial derivatives are crucial in understanding how functions change as multiple variables vary. In contrast to ordinary derivatives, which deal with functions of a single variable, partial derivatives apply to multivariable functions. Let's visualize this with a simple explanation: Imagine you have a surface, like a hill in the landscape. The height of this hill, say \( w \), depends on two input variables, such as your east-west position \( x \) and north-south position \( y \). Now, if you fix your north-south position and only vary your east-west position, the rate at which the height changes is the partial derivative with respect to \( x \), denoted \( \frac{\partial w}{\partial x} \).In our exercise, we calculated several partial derivatives like \( \frac{\partial w}{\partial x} \), \( \frac{\partial w}{\partial y} \), and \( \frac{\partial w}{\partial z} \). These derivatives are essential when we use the chain rule to find how \( w \) changes with respect to another variable, in this case, \( t \). Understanding partial derivatives is a powerful tool when dealing with functions that rely on multiple variables.

Multivariable Functions

Multivariable functions are functions that depend on more than one variable. They can be thought of as functions that take a point in space, specified by several coordinates, and assign a value to it. For instance, consider a weather map where temperature varies with latitude, longitude, and altitude. This temperature function is a multivariable function. In the given exercise, the function \( w = e^{xy+z} \) is a multivariable function, taking three variables \( x \), \( y \), and \( z \) as input. Each of these variables, in turn, depends on two other variables \( s \) and \( t \). This creates an intricate web of dependencies. When differentiating multivariable functions, especially when one seeks to understand how the function changes with respect to one variable while keeping others constant, tools like partial derivatives and the chain rule are invaluable. By unraveling these dependencies, we can pierce through the complexity and find meaningful relationships, just as we did by finding \( \frac{\partial w}{\partial t} \).

Exponential Functions

Exponential functions are common in mathematics and have the form \( e^x \), where \( e \) is a constant approximately equal to 2.71828. This function grows rapidly and is crucial in many fields, from natural sciences to finance. In the essence of calculus, exponential functions have a unique property: they are their own derivatives. That means the rate of change of an exponential function is proportional to the value of the function itself. In our solution, the function \( w = e^{xy + z} \) showcases this exponential behavior. When calculating the partial derivatives, we leverage the property that the derivative of \( e^u \) with respect to \( u \) is \( e^u \). This characteristic simplifies our calculations and allows us to express changes in the function in terms of the exponential itself.Understanding exponential functions is vital as they naturally model processes like growth and decay. In chain rule applications, they elegantly handle dependencies among variables, contributing to comprehensible and manageable solutions.