Question

Given \(z=x e^{-y}, x=\cosh t, y=\cos t\), find \(d z / d t\)

Short answer

Find \(\frac{d z}{d t} = e^{-\cos t}(\sinh t + \cosh t \sin t)\)

Step-by-step solution

- Understand the Chain Rule

To find \(\frac{d z}{d t}\), the chain rule for partial derivatives will be used. This requires calculating partial derivatives of z with respect to both x and y, and then to apply the chain rule combining them with the derivatives of x and y with respect to t.

- Compute Partial Derivative of z with respect to x

For \(\frac{\partial z}{\partial x}\), treat y as a constant: \[\frac{\partial z}{\partial x} = e^{-y}\]

- Compute Partial Derivative of z with respect to y

For \(\frac{\partial z}{\partial y}\), treat x as a constant: \[\frac{\partial z}{\partial y} = -x e^{-y}\]

- Compute derivatives of x and y with respect to t

Calculate \(\frac{d x}{d t}\) and \(\frac{d y}{d t}\): \[\frac{d x}{d t} = \sinh t\] and \[\frac{d y}{d t} = -\sin t\]

- Apply the Chain Rule

Use the chain rule to combine the partial derivatives: \[\frac{d z}{d t} = \frac{\partial z}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial z}{\partial y} \cdot \frac{d y}{d t}\] Substituting the derivatives found in previous steps: \[\frac{d z}{d t} = e^{-y} \cdot \sinh t + (-x e^{-y}) \cdot (-\sin t)\]

- Simplify the Expression

Combine the terms and substitute back the expressions for x and y: \[\frac{d z}{d t} = e^{-\cos t} \sinh t + e^{-\cos t} \cosh t \sin t = e^{-\cos t}(\sinh t + \cosh t \sin t)\]

Key concepts

Partial Derivatives

Partial derivatives play a crucial role in multivariable calculus. Unlike regular derivatives, which measure how a function changes as one single variable changes, partial derivatives measure how a function changes as one specific variable changes while keeping other variables constant.
In the given exercise, we found the partial derivatives of the function \(z = x e^{-y}\) with respect to \(x\) and \(y\). This involved treating the other variable as a constant during differentiation:

• For \(\frac{\partial z}{\partial x}\), we treated \(y\) as constant, yielding \(e^{-y}\).
• For \(\frac{\partial z}{\partial y}\), we treated \(x\) as constant, yielding \(-x e^{-y}\).

These partial derivatives are essential for using the chain rule when differentiating functions of multiple variables.

Differentiation with Respect to a Parameter

In many problems, functions depend on a parameter that can change over time or another variable. Differentiating with respect to this parameter is crucial.

In the exercise, we needed to find \(\frac{d z}{d t}\). Here, \(z=x e^{-y}\) depended on another variable, \(t\), through the functions \(x=\cosh t\) and \(y=\cos t\). Thus, we had to express how both \(x\) and \(y\) change with respect to \(t\) by computing their derivatives:

• The derivative \(\frac{d x}{d t}\) led to \(\sinh t\).
• The derivative \(\frac{d y}{d t}\) led to \(-\sin t\).

By using these derivatives in combination with the chain rule, we were able to find \(\frac{d z}{d t}\).

Exponential Functions

Exponential functions are a fundamental part of calculus. These functions have the form \(e^{u}\), where \(e\) is Euler's number (approximately 2.71828), and \(u\) is some expression or variable.
In the given exercise, the function \(z\) contains the exponential term \(e^{-y}\). Exponential functions have important properties that make differentiation straightforward:

• The derivative of \(e^{u}\) is \(u'e^{u}\) where \(u'\) is the derivative of \(u\) with respect to the variable of interest.
• Exponential functions are always positive.

In our solution, computing the partial derivatives, as well as simplifying \(\frac{dz}{dt}\), relied heavily on understanding these properties.

Hyperbolic Functions

Hyperbolic functions are analogs of trigonometric functions but relate to hyperbolas instead of circles. The most common hyperbolic functions are \(\cosh(t)\) and \(\sinh(t)\). They have important properties and derivatives:

• \(\cosh(t)\) is defined as \((e^{t} + e^{-t})/2\).
• \(\sinh(t)\) is defined as \((e^{t} - e^{-t})/2\).
• The derivative of \(\cosh(t)\) is \(\sinh(t)\).
• The derivative of \(\sinh(t)\) is \(\cosh(t)\).

In our exercise, we used these identities to find that \(\frac{d x}{d t} = \sinh t\) and combined these results to apply the chain rule effectively in differentiating \(z\).