Question

Find the first partial derivatives with respect to \(x, y,\) and \(z\). $$ F(x, y, z)=\ln \sqrt{x^{2}+y^{2}+z^{2}} $$

Short answer

The derivatives are \(\frac{\partial F}{\partial x} = \frac{x}{x^{2} + y^{2}+ z^{2}}\), \(\frac{\partial F}{\partial y} = \frac{y}{x^{2} + y^{2}+ z^{2}}\), \(\frac{\partial F}{\partial z} = \frac{z}{x^{2} + y^{2}+ z^{2}}\).

Step-by-step solution

Simplify the function

Rewrite the given function as \(F(x, y, z) = \frac{1}{2} \ln(x^{2} + y^{2} + z^{2})\). This step simplifies the function and will make the derivative easier to calculate.

Carry out partial differentiation with respect to \(x\)

The partial derivative with respect to \(x\) is given by the chain rule as \(\frac{\partial F}{\partial x} = \frac{1}{2} * \frac{1}{(x^{2}+y^{2}+z^{2})} * 2x\), simplifies to \(\frac{x}{x^{2} + y^{2}+ z^{2}}\).

Carry out partial differentiation with respect to \(y\)

The partial derivative with respect to \(y\) similar to \(x\), which is \(\frac{\partial F}{\partial y} = \frac{1}{2} * \frac{1}{(x^{2}+y^{2}+z^{2})} * 2y\), simplifies to \(\frac{y}{x^{2} + y^{2}+ z^{2}}\).

Carry out partial differentiation with respect to \(z\)

The partial derivative with respect to \(z\) is given by \(\frac{\partial F}{\partial z} = \frac{1}{2} * \frac{1}{(x^{2}+y^{2}+z^{2})} * 2z\), simplifies to \(\frac{z}{x^{2} + y^{2}+ z^{2}}\).

Key concepts

Multivariable Calculus

Multivariable calculus is an extension of single-variable calculus that deals with functions of several variables. It involves concepts such as partial derivatives and multiple integrals. In problems like our original exercise, instead of differentiating with respect to a single variable, we differentiate with respect to multiple variables such as x, y, and z.

When working with multivariable functions like F(x, y, z), it's important to understand the geometric interpretation: each partial derivative represents the rate of change of F along the axis of the variable in question while keeping the others constant. Think of it as slicing the function's graph along one of the axes and seeing how steeply the curve rises or falls.

Chain Rule

The chain rule is a fundamental principle in calculus that is used to compute the derivative of a composite function. When dealing with multivariable functions, the chain rule helps us differentiate a function with respect to one variable while considering the impact of the other variables.

For example, when finding the partial derivative of F(x, y, z) with respect to x, we consider y and z as constants and only differentiate x's terms. The chain rule allows us to break down the differentiating of complicated functions into simpler parts, which can then be calculated step by step as seen in the exercise solution.

Partial Differentiation

Partial differentiation, one of the key operations in multivariable calculus, deals with the derivatives of functions of multiple variables with respect to one variable at a time. It’s like asking the question 'How does the function change if I tweak just this one variable, while keeping others fixed?'

In our exercise, the process of simplifying F(x, y, z) before differentiation is crucial. It transforms the radical into a logarithmic form that is easier to differentiate. Then, finding the partial derivative with respect to each variable demonstrates the multidimensional rate of change of the function. Always remember that each partial derivative is calculated independently of the others.

Logarithmic Functions

Logarithmic functions are inverses of exponential functions, and they come up frequently in calculus, especially in growth and decay problems. In the context of differentiation, the derivatives of logarithmic functions have their own set of rules, distinct from those for algebraic or trigonometric functions.

The function F(x, y, z) in our exercise involves the natural logarithm of a square root, a composite function. Simplifying the square root into a power of 1/2 allows us to apply the logarithm power rule before differentiation, making the partial derivatives easier to find. This highlights the importance of algebraic manipulation in calculus to make complex problems more tractable.