Question

$$\text { If } z=x e^{-y} \text { and } x=\cosh t, y=\cos s, \text { find } \partial z / \partial s \text { and } \partial z / \partial t$$.

Short answer

\( \partial z / \partial s = \cosh(t) e^{-\cos(s)} \sin(s) \) and \( \partial z / \partial t = e^{-\cos(s)} \sinh(t) \).

Step-by-step solution

- Express z in terms of t and s

Given that \( z = x e^{-y} \), \( x = \cosh(t) \), and \( y = \cos(s) \). Substitute \( x \) and \( y \) into the expression for \( z \): \( z = \cosh(t)\cdot e^{-\cos(s)} \).

- Find \( \partial z / \partial s \)

To find the partial derivative of \( z \) with respect to \( s \), use the chain rule. Start by differentiating \( z = \cosh(t) \cdot e^{-\cos(s)} \) with respect to \( s \): \( \frac{\partial z}{\partial s} = \cosh(t) \cdot \frac{d}{ds}\left( e^{-\cos(s)}\right) \). The derivative of \( e^{-\cos(s)} \) with respect to \( s \) is \( e^{-\cos(s)} \cdot \sin(s) \) considering the chain rule. Hence, \( \frac{\partial z}{\partial s} = \cosh(t) \cdot e^{-\cos(s)} \cdot \sin(s) \).

- Find \( \partial z / \partial t \)

To find the partial derivative of \( z \) with respect to \( t \): Differentiate \( z = \cosh(t) \cdot e^{-\cos(s)} \) with respect to \( t \): \( \frac{\partial z}{\partial t} = e^{-\cos(s)} \cdot \frac{d}{dt}(\cosh(t)) \). The derivative of \( \cosh(t) \) with respect to \( t \) is \( \sinh(t) \). Hence, \( \frac{\partial z}{\partial t} = e^{-\cos(s)} \cdot \sinh(t) \).

Key concepts

Chain Rule

The chain rule is an essential tool in calculus when dealing with composite functions. It allows us to take the derivative of a function that depends on another function. Here's how it works: If a variable z depends on y, which in turn depends on x, the chainrule helps us differentiate z with respect to x.

In simpler terms:
If \(z = f(y)\) and \(y = g(x)\), then the chain rule states:
\[ \frac{\partial z}{\partial x} = \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial x} \]
The key is to first differentiate \(z\) with respect to \(y\) and then multiply it by the derivative of \(y\) with respect to \(x\).

In the given exercise, we used the chain rule to find how \(z\) changes with respect to \(s\) and \(t\). When finding \(\frac{\partial z}{\partial s}\), we treated \(t\) as a constant and vice versa. This method simplifies the differentiation process and is crucial for tackling more complex problems.

Hyperbolic Functions

Hyperbolic functions can be thought of as the 'sibling' of trigonometric functions but for the hyperbola instead of the circle. These functions include \(\text{cosh}(x)\) and \(\text{sinh}(x)\). They have similar properties to cosine and sine functions, which makes them integral in certain physics and engineering problems.

For hyperbolic cosine:\ \ \[ \cosh(x) = \frac{e^x + e^{-x}}{2} \]
For hyperbolic sine:\ \ \[ \sinh(x) = \frac{e^x - e^{-x}}{2} \]
In our exercise, \(x\) was given as \(\text{cosh}(t)\). When differentiating \(\text{cosh}(t)\) with respect to \(t\), we used the definition of the hyperbolic functions:
\ \ \[ \frac{d}{dt} \cosh(t) = \sinh(t) \]
These hyperbolic functions are pivotal because they often appear in solutions to differential equations describing physical phenomena like heat conduction and wave propagation.

Exponential Functions

Exponential functions describe a constant rate of growth or decay. For example, \(e^x\) grows rapidly as \(x\) increases and decays to zero as \(x\) decreases.
One of the key properties of exponential functions is the fact that their derivatives are proportional to themselves. This means:
\ \ \[ \frac{d}{dx} (e^x) = e^x \]

In the context of the exercise, this property became essential when dealing with \(e^{-\text{cos}(s)}\). To differentiate this, we used the chain rule. The negative sign inside the exponent and the cosine function added complexity, but understanding the basic property of \(e^x\) allowed us to handle it with ease.
\ \ \[ \frac{d}{ds}(e^{-\text{cos}(s)}) = e^{-\text{cos}(s)} \cdot (-\sin(s)) \]
This step involves both differentiation of exponential functions and knowledge of trigonometric identities. Thus, the familiarity with these foundational functions simplifies tackling complex derivatives as seen in this problem.