Question

Find all the second partial derivatives in each of the following cases: (a) \(z=x \sin y(\) b) \(z=y \cos x\) (c) \(z=y \mathrm{e}^{2 x}\left(\right.\) d) \(z=y \mathrm{e}^{-x}\)

Short answer

Question: Find all the second partial derivatives for the following functions: (a) \(z = x \sin{y}\) (b) \(z = y \cos{x}\) (c) \(z = y e^{2x}\) (d) \(z = y e^{-x}\) Answer: (a) \(\frac{\partial^2 z}{\partial x^2} = 0\), \(\frac{\partial^2 z}{\partial x \partial y} = \cos{y}\), \(\frac{\partial^2 z}{\partial y^2} = -x \sin{y}\) (b) \(\frac{\partial^2 z}{\partial x^2} = -y \cos{x}\), \(\frac{\partial^2 z}{\partial x \partial y} = -\sin{x}\), \(\frac{\partial^2 z}{\partial y^2} = 0\) (c) \(\frac{\partial^2 z}{\partial x^2} = 4y e^{2x}\), \(\frac{\partial^2 z}{\partial x \partial y} = 2 e^{2x}\), \(\frac{\partial^2 z}{\partial y^2} = 0\) (d) \(\frac{\partial^2 z}{\partial x^2} = y e^{-x}\), \(\frac{\partial^2 z}{\partial x \partial y} = - e^{-x}\), \(\frac{\partial^2 z}{\partial y^2} = 0\)

Step-by-step solution

(a) First partial derivatives of \(z = x \sin{y}\)

First, we will find the first partial derivatives of the given function \(z = x \sin{y}\) with respect to \(x\) and \(y\). \(\frac{\partial z}{\partial x} = \sin{y}\) \(\frac{\partial z}{\partial y} = x \cos{y}\)

(a) Second partial derivatives of \(z = x \sin{y}\)

Now, we will find the second partial derivatives by differentiating the first partial derivatives w.r.t \(x\) and \(y\). \(\frac{\partial^2 z}{\partial x^2} = 0\) \(\frac{\partial^2 z}{\partial x \partial y} = \cos{y}\) \(\frac{\partial^2 z}{\partial y^2} = -x \sin{y}\)

(b) First partial derivatives of \(z = y \cos{x}\)

First, we will find the first partial derivatives of the given function \(z = y \cos{x}\) with respect to \(x\) and \(y\). \(\frac{\partial z}{\partial x} = -y \sin{x}\) \(\frac{\partial z}{\partial y} = \cos{x}\)

(b) Second partial derivatives of \(z = y \cos{x}\)

Now, we will find the second partial derivatives by differentiating the first partial derivatives w.r.t \(x\) and \(y\). \(\frac{\partial^2 z}{\partial x^2} = -y \cos{x}\) \(\frac{\partial^2 z}{\partial x \partial y} = -\sin{x}\) \(\frac{\partial^2 z}{\partial y^2} = 0\)

(c) First partial derivatives of \(z = y e^{2x}\)

First, we will find the first partial derivatives of the given function \(z = y e^{2x}\) with respect to \(x\) and \(y\). \(\frac{\partial z}{\partial x} = 2y e^{2x}\) \(\frac{\partial z}{\partial y} = e^{2x}\)

(c) Second partial derivatives of \(z = y e^{2x}\)

Now, we will find the second partial derivatives by differentiating the first partial derivatives w.r.t \(x\) and \(y\). \(\frac{\partial^2 z}{\partial x^2} = 4y e^{2x}\) \(\frac{\partial^2 z}{\partial x \partial y} = 2 e^{2x}\) \(\frac{\partial^2 z}{\partial y^2} = 0\)

(d) First partial derivatives of \(z = y e^{-x}\)

First, we will find the first partial derivatives of the given function \(z = y e^{-x}\) with respect to \(x\) and \(y\). \(\frac{\partial z}{\partial x} = -y e^{-x}\) \(\frac{\partial z}{\partial y} = e^{-x}\)

(d) Second partial derivatives of \(z = y e^{-x}\)

Now, we will find the second partial derivatives by differentiating the first partial derivatives w.r.t \(x\) and \(y\). \(\frac{\partial^2 z}{\partial x^2} = y e^{-x}\) \(\frac{\partial^2 z}{\partial x \partial y} = - e^{-x}\) \(\frac{\partial^2 z}{\partial y^2} = 0\) Now we have calculated all the second partial derivatives for each of the given functions.

Key concepts

Partial Differentiation

Partial differentiation is a foundational concept in multivariable calculus, crucial for analyzing functions of several variables. It involves taking the derivative of a function with respect to one variable at a time while treating all other variables as constants.

For example, let's consider a function with two variables, such as the function from the exercise: \( z = x \sin{y} \). To find the partial derivative of \( z \) with respect to \( x \), we differentiate \( z \) with respect to \( x \), treating \( y \) as a constant. The result is \( \frac{\partial z}{\partial x} = \sin{y} \), since \( x \) is the variable of differentiation and \( \sin{y} \) is treated as a constant factor. Similarly, when differentiating with respect to \( y \), the value \( x \) is considered constant, resulting in \( \frac{\partial z}{\partial y} = x \cos{y} \).

Understanding partial differentiation is essential for solving more complex problems in multivariable calculus, where the interactions between variables can have significant implications.

Multivariable Calculus

Multivariable calculus extends the principles of one-variable calculus to functions with multiple variables. This branch of calculus allows for the exploration of surfaces, curves, and spaces that cannot be described by a single-variable function.

In the context of the exercise provided, we see an application of multivariable calculus principles where we differentiate functions like \( z = y \cos{x} \) and \( z = y e^{2x} \) with respect to each variable. By finding the first and second partial derivatives, we describe how the function's surface responds to changes in each direction individually. These derivatives give us insight into the function's behavior, such as its rates of change, curvature, and potential maximum or minimum points.

Through multivariable calculus, we can solve problems in physics, engineering, and economics, dealing with phenomena that change in more than one dimension.

Chain Rule in Calculus

The chain rule is a fundamental technique in calculus for differentiating compositions of functions. It allows us to compute the derivative of a function that is built from other functions whose derivatives are known.

In the context of multivariable calculus and partial differentiation, the chain rule becomes even more nuanced. If a function \( z \) depends on multiple variables which themselves depend on another variable, say \( t \), then the chain rule provides a way to express \( \frac{dz}{dt} \) in terms of partial derivatives. For instance, if \( z \) is expressed as \( z(x(t), y(t)) \), then using the multivariable chain rule, \( \frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt} \).

This procedure is critical when dealing with dynamic systems where one quantity's change depends on multiple interrelated variables.

First Partial Derivatives

First partial derivatives represent the rate at which a function changes as one variable is varied, while others are held constant. They are the first step in understanding a function's behavior in multivariable spaces. The computation of first partial derivatives was demonstrated in the solution to the exercise as the initial derivatives of \( z \) with respect to each independent variable.

For example, the first partial derivatives of the function \( z = y e^{2x} \) with respect to \( x \) and \( y \) were found to be \( \frac{\partial z}{\partial x} = 2ye^{2x} \) and \( \frac{\partial z}{\partial y} = e^{2x} \), respectively. These derivatives signify how the function escalates or descends in the direction of each variable. In practical scenarios, understanding first partial derivatives can facilitate the optimization of processes or the study of changes, such as temperature gradients, economic growth rates, and more.