Problem 4 - College Algebra and Calculus: An Applied Approach

Question

Find the first partial derivatives with respect to \(x\) and with respect to \(y\). $$ f(x, y)=x+4 y^{3 / 2} $$

Step-by-Step Solution

Answer

The first partial derivative of \(f(x, y) = x + 4 y^{3/2}\) with respect to \(x\) is 1, and with respect to \(y\) is \(6 y^{1/2}.\)

Step 1: Compute the partial derivative with respect to \(x\)

To compute the partial derivative of \(f(x, y) = x + 4 y^{3/2}\) with respect to \(x\), treat \(y\) as a constant. The derivative of a constant is zero, and the derivative of \(x\) with respect to \(x\) is one. This gives \(\frac{\partial f}{\partial x} = 1 + 0 = 1.\)

Step 2: Compute the partial derivative with respect to \(y\)

To compute the partial derivative of \(f(x, y) = x + 4 y^{3/2}\) with respect to \(y\), treat \(x\) as a constant. The derivative of a constant is zero, and the derivative of \(y^{3/2}\) with respect to \(y\) is \(\frac{3}{2} y^{1/2}\). Then multiply this by the constant multiple 4 to get \(\frac{\partial f}{\partial y} = 0 + 4*\frac{3}{2} y^{1/2} = 6 y^{1/2}.\)

Key Concepts

Multivariable Calculus

Multivariable Calculus is an extension of single-variable calculus to functions of several variables. While single-variable calculus involves the study of functions with one input, multivariable calculus analyzes functions that take two or more inputs. In multivariable calculus, we often deal with functions of the form \( f(x, y) \), where \( x \) and \( y \) are independent variables. Instead of a single derivative, we explore partial derivatives—derivatives with respect to one variable while holding others constant. This approach helps us understand how changes in one variable affect the function's output, which is essential in many fields such as physics, engineering, and economics. One common question in multivariable calculus is finding these partial derivatives to understand the behavior of functions more thoroughly.

Derivative Rules

Derivative rules are essential tools that simplify the differentiation process. In the context of multivariable calculus, these rules adapt to accommodate partial derivatives. When computing partial derivatives:

• Treat all variables, except the one you're differentiating with respect to, as constants. This is crucial for isolating the effect of one variable at a time.
• Apply basic differentiation rules of power, constants, and sums. For example, the derivative of \(x\) with respect to \(x\) is always 1, and for any constant, it is 0.

To illustrate, in the function \( f(x, y) = x + 4y^{3/2} \), when finding the partial derivative with respect to \(x\), you treat \(y\) as a constant. Similarly, when differentiating with respect to \(y\), treat \(x\) as a constant.This simplifies the problem significantly, allowing us to apply our knowledge of single-variable derivatives in a new context.

Differentiation Techniques

Differentiation techniques extend beyond basic power rules, especially in multivariable contexts. A key technique in this area is identifying which terms to differentiate and treating other variables as constants. For example, in the function \( f(x, y) = x + 4y^{3/2} \), identifying that \(x\) and \(4y^{3/2}\) are independent terms enables efficient differentiation.

• To differentiate \( f(x, y) \) with respect to \(x\), recognize that \(4y^{3/2}\) is constant with respect to \(x\), simplifying the computation to a straightforward derivative of \(x\).
• To differentiate with respect to \(y\), treat \(x\) as a constant and apply the power rule to \(y^{3/2}\). The power rule here states that \( \frac{d}{dy}[y^n] = ny^{n-1} \), which results in \( \frac{3}{2}y^{1/2} \).

Employing these techniques makes handling functions with more than one variable manageable, aiding in clear understanding and problem-solving.