Question

Compute the partial derivatives \((\partial f / \partial x)_{y}\) and \((\partial f / \partial y)_{x}\) for the following functions: (a) \(f(x, y)=3 x^{2}+y^{5}\) (b) \(f(x, y)=x^{10} y^{1 / 2}\) (c) \(f(x, y)=x+y^{2}+3\) (d) \(f(x, y)=5 x\)

Short answer

Partial derivatives: (a) w.r.t x: 6x, w.r.t y: 5y^4. (b) w.r.t x: 10x^9 y^{1/2}, w.r.t y: (1/2)x^{10} y^{-1/2}. (c) w.r.t x: 1, w.r.t y: 2y. (d) w.r.t x: 5, w.r.t y: 0.

Step-by-step solution

Define the function and the derivatives

To find the partial derivatives, first define the function. Then, find the partial derivative with respect to x, keeping y constant, and the partial derivative with respect to y, keeping x constant.

Calculate partial derivatives for (a)

Given function: \( f(x, y) = 3x^2 + y^5 \)\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(3x^2 + y^5) = 3 \frac{d}{dx}(x^2) = 6x \]\[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(3x^2 + y^5) = 5y^4 \]

Calculate partial derivatives for (b)

Given function: \( f(x, y) = x^{10} y^{1/2} \)\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x^{10} y^{1/2}) = 10x^9 y^{1/2} \]\[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x^{10} y^{1/2}) = x^{10}(1/2)y^{-1/2} = \frac{1}{2}x^{10} y^{-1/2} \]

Calculate partial derivatives for (c)

Given function: \( f(x, y) = x + y^2 + 3 \)\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(x + y^2 + 3) = 1 \]\[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(x + y^2 + 3) = 2y \]

Calculate partial derivatives for (d)

Given function: \( f(x, y) = 5x \)\[ \frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(5x) = 5 \]\[ \frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(5x) = 0 \]

Key concepts

calculus

Calculus is a branch of mathematics focused on studying rates of change (differentiation) and accumulation of quantities (integration). The core ideas of calculus can be divided into differential calculus, which deals with the concept of a derivative, and integral calculus, which is concerned with computing the area underneath curves.

In this problem, we focus on differential calculus, specifically on finding the partial derivatives of functions involving two variables. The derivative gives us the rate at which a function changes as its input changes. Here, we differentiate with respect to one variable at a time, keeping the other variable constant. This process is crucial in multivariable calculus to understand how changes in one variable affect a function when others are held fixed.

multivariable calculus

Multivariable calculus extends the principles of calculus to functions of several variables. It is essential for understanding physical phenomena involving more than one changing parameter.

In this specific problem, we deal with functions of two variables, say, x and y. Calculating partial derivatives helps us understand how each variable individually affects the function. For instance, given a function like f(x, y) = 3x^2 + y^5, we compute \(\frac{\frac{\text{partial } f}{\text{partial } x}}{y}\) by treating y as a constant and differentiating with respect to x, and vice versa for \(\frac{\frac{\text{partial } f}{\text{partial } y}}{x}\).

Neither x nor y stands alone when analyzing the relationships and rates of change involving several dimensions. This concept becomes particularly powerful in fields such as physics, economics, and engineering, where system behaviors depend on multiple variables.

derivatives

Derivatives are a fundamental concept in calculus that measure how a function changes as its input changes. For single-variable functions, this is expressed as dy/dx. For multivariable functions, we use partial derivatives, denoted \(\frac{\frac{\text{partial } f}{\text{partial } x}}{y}\) and \(\frac{\frac{\text{partial } f}{\text{partial } y}}{x}\).

To compute partial derivatives, we follow these steps:

• Identify the function.
• Differentiating with respect to one variable, keeping the other variables constant.

Let's break down the solutions from the exercise:
For function (a) \( f(x, y) = 3x^2 + y^5 \),
\(\frac{\frac{\text{partial } f}{\text{partial } x}}{y} = 6x\) because \(\frac{d(3x^2)}{dx} = 6x\) and \( f(x, y) = 6x \) when y is constant. For the same function, \(\frac{\frac{\text{partial } f}{\text{partial } y}}{x} = 5y^4\) because \(\frac{d(y^5)}{dy} = 5y^4\).

By carefully understanding these calculating steps on a variety of function types, students grasp how each variable proportionally impacts the overall function. This deeper insight is indispensable for applying calculus to real-world problems.