Problem 121 - Calculus Volume 3

Question

For the following exercises, calculate the partial derivatives. \(\quad \frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) for \(z=\ln \left(x^{6}+y^{4}\right)\).

Step-by-Step Solution

Answer

\( \frac{\partial z}{\partial x} = \frac{6x^5}{x^6+y^4} \) and \( \frac{\partial z}{\partial y} = \frac{4y^3}{x^6+y^4} \).

Step 1: Identify the Function

The function given is \( z = \ln(x^6 + y^4) \). We need to find the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) for this function.

Step 2: Differentiate with Respect to x

To find \( \frac{\partial z}{\partial x} \), we treat \( y \) as a constant. First, differentiate \( \ln(u) \) with respect to \( u \), where \( u = x^6 + y^4 \). The derivative is \( \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( \frac{du}{dx} = 6x^5 \). Thus, \( \frac{\partial z}{\partial x} = \frac{1}{x^6 + y^4} \cdot 6x^5 = \frac{6x^5}{x^6 + y^4} \).

Step 3: Differentiate with Respect to y

To find \( \frac{\partial z}{\partial y} \), treat \( x \) as a constant. Using the same process, differentiate \( \ln(u) \) with respect to \( y \). Here, \( \frac{du}{dy} = 4y^3 \). Thus, \( \frac{\partial z}{\partial y} = \frac{1}{x^6 + y^4} \cdot 4y^3 = \frac{4y^3}{x^6 + y^4} \).

Key Concepts

Calculus

Calculus is a vast field in mathematics that deals with change and motion. It provides tools and techniques to analyze systems that change dynamically. One of the most important components of calculus is differentiation, which allows us to understand how a function changes at any point. In the context of functions involving more than one variable, such as the logarithmic function given in the exercise, calculus aids in computing partial derivatives, which are used to determine how the function changes with respect to one variable while keeping other variables constant. Understanding calculus is crucial because it opens the door to solving real-world problems involving motion, growth, and decay. By learning calculus, you gain insights into numerous applications in fields like physics, engineering, and economics.

Natural Logarithm

The natural logarithm, denoted as \(ln(x)\), is a fundamental mathematical function. It is the inverse of the exponential function \(e^x\). The base of the natural logarithm is \(e\), an irrational constant approximately equal to 2.71828. Natural logarithms are widely used in calculus because they have unique differentiation properties. For instance, the derivative of \(ln(u)\) is \(\frac{1}{u} \frac{du}{dx}\), making it essential for tackling logarithmic differentiation problems. This property was used to solve the exercise where the function \(z = \ln(x^6 + y^4)\) was differentiated.Understanding natural logarithms is crucial, particularly in fields like biology and chemistry, where exponential growth and decay rates are modeled.

Differentiation

Differentiation is the process of finding the derivative of a function, which indicates the rate at which a function is changing at any point. In simpler terms, it's like asking how steep a hill is while climbing it. In the exercise, differentiation was performed to find the partial derivatives \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\). The steps involved applying the chain rule to the natural logarithm function \(z = \ln(x^6 + y^4)\). By treating one variable as constant and differentiating with respect to the other, we found how the function changes in each direction.Differentiation is a core concept in calculus, helping to solve problems across different domains by providing a mathematical framework to analyze change.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions with more than one variable. Unlike single-variable calculus, where we only consider functions such as \(f(x)\), multivariable calculus explores functions like \(f(x, y)\), allowing us to study surfaces and more complex systems. Partial derivatives are vital in this area as they measure how a function changes with respect to individual variables. In the exercise, calculating \(\frac{\partial z}{\partial x}\) and \(\frac{\partial z}{\partial y}\) required treating the other variable as a constant and focusing on one variable at a time. This method is extremely useful in fields like engineering and physics, where multiple variables often interact.Mastering multivariable calculus unlocks new ways to model and solve intricate problems, making it an invaluable skill in many scientific areas.