Question

Short-answer questions. (a) Compute the partial derivatives: \((\partial f / \partial x) y\) and \((\partial f / \partial y)_{x}\) for the following functions: $$ f(x, y)=\ln (2 x)+5 y^{3}, $$

Short answer

\( \partial f / \partial x = \frac{1}{x} \) and \( \partial f / \partial y = 15y^2 \)

Step-by-step solution

- Identify the function

The given function is \( f(x, y) = \ln(2x) + 5y^3 \). Break it into two parts: \( \ln(2x) \) and \( 5y^3 \).

- Compute \( \partial f / \partial x \)

We need to find the partial derivative of \( f \) with respect to \(x\). Use the chain rule: \( \partial (\ln(2x)) / \partial x = \frac{1}{2x} \cdot 2 = \frac{1}{x} \). The term \( 5y^3 \) is independent of \(x\) and becomes zero. So, \( \partial f / \partial x = \frac{1}{x} \).

- Compute \( \partial f / \partial y \)

We need to find the partial derivative of \( f \) with respect to \(y\). The term \( \ln(2x) \) is independent of \(y\) and becomes zero. Using the power rule: \( \partial (5y^3) / \partial y = 3 \cdot 5y^2 = 15y^2 \). So, \( \partial f / \partial y = 15y^2 \).

Key concepts

chain rule

The chain rule is a fundamental concept in calculus, especially in the context of functions of several variables. To put it simply, the chain rule is used to compute the derivative of a composite function. This means you're looking at how one function changes with respect to another.

For example, if you have a function composed of two functions, say \(f(x) = g(h(x))\), the chain rule allows you to differentiate it as:\[\frac{df}{dx} = g'(h(x)) \,\cdot\, h'(x)\].

In the realm of partial derivatives, the chain rule helps us differentiate functions where variables are implicitly dependent on each other. This is crucial when dealing with multivariable functions like \(f(x, y) = \ln(2x) + 5y^3\). The chain rule becomes handy, especially while taking the derivative of \(\ln(2x)\) with respect to \(x\).

Understanding the chain rule requires practice. Here are a few key steps you can follow:

• Identify the outer function and the inner function.
• Differentiate the outer function while keeping the inner function unchanged.
• Multiply by the derivative of the inner function.

Mastering these steps can greatly simplify working with more complicated functions.

power rule

The power rule is one of the most straightforward differentiation rules you'll encounter in calculus. It states that if you have a function of the form \(f(x) = x^n\), where \(n\) is any real number, then the derivative of the function is \(f'(x) = nx^{n-1}\).

In simpler terms, you bring down the exponent as a coefficient in front and then subtract one from the original exponent.

For example, consider the partial derivative \(\partial f/\partial y\) of the function \(f(x, y) = \ln(2x) + 5y^3\). The term \(5y^3\) can be differentiated using the power rule:\[ \frac{\partial}{\partial y}(5y^3) = 3 \cdot 5y^{3-1} = 15y^2 \].

Because the term \(\ln(2x)\) doesn't involve \(y\), its partial derivative with respect to \(y\) is zero. Combining these results, we find \(\partial f/\partial y = 15y^2\).

This is a prime example of the power rule in action! It simplifies the process of finding derivatives by giving us a straightforward means to tackle polynomial terms.

multivariable calculus

Multivariable calculus extends the principles of calculus to functions of several variables. This area of mathematics allows us to explore how functions change when they depend on more than one input.

Take the function \(f(x, y) = \ln(2x) + 5y^3\) for instance. This is a two-variable function, where both \(x\) and \(y\) influence the outcome. When dealing with such functions, partial derivatives play a key role.

A partial derivative measures the rate of change of the function with respect to one variable while keeping the other variables constant. By isolating how the function changes in one direction, we gain deeper insights into its behavior.

• For \(\partial f / \partial x\), we treat \(y\) as a constant and differentiate \(f(x, y)\) solely with respect to \(x\).
• For \(\partial f / \partial y\), we treat \(x\) as a constant and differentiate \(f(x, y)\) solely with respect to \(y\).

In our given example, we found:

• \(\partial f / \partial x = \frac{1}{x}\)
• \(\partial f / \partial y = 15y^2\)

These derivatives tell us how the function \(f(x, y)\) changes as \(x\) or \(y\) changes, providing a powerful tool for analyzing multivariable systems.