Question

Find the first partial derivatives with respect to \(x, y,\) and \(z\). $$ G(x, y, z)=\frac{1}{\sqrt{1-x^{2}-y^{2}-z^{2}}} $$

Short answer

The first partial derivatives of the function \(G(x, y, z)=\frac{1}{\sqrt{1-x^{2}-y^{2}-z^{2}}}\) with respect to \(x, y, z\) are \(\frac{\partial G}{\partial x} = x(1-x^{2}-y^{2}-z^{2})^{-3/2}\), \(\frac{\partial G}{\partial y} = y(1-x^{2}-y^{2}-z^{2})^{-3/2}\) and \(\frac{\partial G}{\partial z} = z(1-x^{2}-y^{2}-z^{2})^{-3/2}\) respectively.

Step-by-step solution

Recognize the Problem Type

This is a standard problem of finding the first partial derivatives of a function with respect to multiple variables. The given function is \(G(x, y, z)=\frac{1}{\sqrt{1-x^{2}-y^{2}-z^{2}}}\)

Apply the Chain Rule

We would apply the chain rule in taking partial derivatives. First, let us rewrite the function more conveniently for applying the chain rule: \(G(x, y, z) = (1-x^{2}-y^{2}-z^{2})^{-1/2}\)

Compute the partial derivative with respect to x

The partial derivative of \(G\) with respect to \(x\) is: \( \frac{\partial G}{\partial x} = -\frac{1}{2}(1-x^{2}-y^{2}-z^{2})^{-3/2}*(-2x) = x(1-x^{2}-y^{2}-z^{2})^{-3/2} \)

Compute the partial derivative with respect to y

The partial derivative of \(G\) with respect to \(y\) is: \( \frac{\partial G}{\partial y} = -\frac{1}{2}(1-x^{2}-y^{2}-z^{2})^{-3/2}*(-2y) = y(1-x^{2}-y^{2}-z^{2})^{-3/2} \)

Compute the partial derivative with respect to z

The partial derivative of \(G\) with respect to \(z\) is: \( \frac{\partial G}{\partial z} = -\frac{1}{2}(1-x^{2}-y^{2}-z^{2})^{-3/2}*(-2z) = z(1-x^{2}-y^{2}-z^{2})^{-3/2} \)

Key concepts

Chain Rule Differentiation

Understanding the chain rule is essential in calculus, especially when dealing with functions composed of other functions. This concept is even more crucial in multivariable calculus, where the relationships between variables can be quite complex.

The chain rule allows us to differentiate compositions of functions by tackling each function separately and then combining the derivatives according to the rule. In the context of partial derivatives, when multiple variables are involved, this rule guides us on how to differentiate with respect to each variable by treating other variables as constants.

For example, given a function like G(x, y, z) = (1-x^2-y^2-z^2)^{-1/2}, when applying the chain rule, we consider the inner function u(x, y, z) = 1-x^2-y^2-z^2 and the outer function f(u) = u^{-1/2}. The chain rule would then imply taking the derivative of the outer function with respect to u, multiplying it by the derivative of u with respect to the variable of interest (x, y, or z).

Practical Implementation of the Chain Rule

By applying the chain rule to our example, we get the partial derivatives with respect to x, y, and z. We treat each variable in isolation, and for each differentiation, we effectively freeze the other variables, regarding them as constants.

The steps provided in the solution are a direct application of the chain rule. By rewriting the function to highlight the composition and taking derivatives step by step, we can find the required partial derivatives.

Multivariable Calculus

Multivariable calculus extends the principles of single variable calculus to functions of multiple variables. This field deals with functions like G(x, y, z) which depend on more than one independent variable. In these cases, the concept of a 'partial derivative' becomes important.

A key idea in multivariable calculus is to understand how a function changes with respect to each variable individually. This is where partial derivatives come into play—they measure the rate of change of the function with respect to one variable, while the other variables are held constant.

Applications of Multivariable Calculus

From optimizing functions in economics to modeling physical phenomena in engineering and science, the applications of multivariable calculus are broad and underpin many advances in technology and analytic studies. For instance, understanding how changing certain parameters affects a given outcome can be crucial in improving complex systems.

As we see in the solution to the exercise, the calculations involve taking derivatives with respect to one variable at a time, which is a foundational skill in multivariable calculus.

First Order Partial Derivatives

First order partial derivatives refer to the derivatives that are taken with respect to one variable at a time in a multivariable function. These derivatives are crucial because they provide immediate information about the function's local behavior in the direction of each independent variable.

In the exercise, the first order partial derivatives of G(x, y, z) indicate how the function moves along the x, y, and z axes, when you slightly change one variable, but keep the others fixed. It's much like examining the slopes in each dimension individually.

Interpreting First Order Partial Derivatives

The calculated first order partial derivatives not only inform us about the slope of G in each direction but also help in identifying the direction and rate of the fastest increase or decrease at a point. For our given function, the signs and magnitudes of these derivatives would indicate the character and geometry of the function near the point of differentiation.

These concepts serve as the building blocks for more advanced topics in multivariable calculus, such as gradient vectors and directional derivatives, significantly adding to our toolkit in analyzing multivariable functions.