Find The First Partial Derivatives Of The Function.

Greels
May 02, 2025 · 5 min read

Table of Contents
Finding the First Partial Derivatives of a Function: A Comprehensive Guide
Partial derivatives are a fundamental concept in multivariable calculus, extending the idea of a derivative to functions of multiple variables. Understanding how to find them is crucial for various applications in physics, engineering, economics, and machine learning. This article provides a comprehensive guide to finding first partial derivatives, covering various techniques and complexities. We’ll explore different types of functions, including those involving implicit differentiation and chain rule applications.
Understanding Partial Derivatives
Before diving into the mechanics of calculating partial derivatives, let's solidify our understanding of what they represent. Imagine a function, f(x, y)
, which maps points in a two-dimensional space (x, y) to a scalar value. A partial derivative measures the rate of change of this function with respect to one variable, holding all other variables constant.
For instance, the partial derivative of f(x, y)
with respect to x
, denoted as ∂f/∂x or f<sub>x</sub>, represents the instantaneous rate of change of f
as x
changes, while y
remains fixed. Similarly, ∂f/∂y or f<sub>y</sub> represents the rate of change of f
with respect to y
, keeping x
constant.
This "holding other variables constant" is crucial. It allows us to analyze the influence of each individual variable on the function's behavior without the confounding effects of other variables.
Calculating First Partial Derivatives: Techniques and Examples
Calculating first partial derivatives involves applying standard differentiation rules, treating all variables except the one being differentiated as constants. Let's explore various examples.
1. Polynomials and Simple Algebraic Functions
These are the easiest to handle. We simply differentiate with respect to the variable of interest, treating other variables as constants.
Example 1:
Find the first partial derivatives of f(x, y) = x² + 3xy + y³
.
-
∂f/∂x: Differentiating with respect to
x
, we treaty
as a constant: ∂f/∂x = 2x + 3y -
∂f/∂y: Differentiating with respect to
y
, we treatx
as a constant: ∂f/∂y = 3x + 3y²
Example 2:
Find the first partial derivatives of g(x, y, z) = x²yz + sin(x) + e<sup>y</sup>
.
-
∂g/∂x: 2xyz + cos(x)
-
∂g/∂y: x²z + e<sup>y</sup>
-
∂g/∂z: x²y
2. Trigonometric Functions
Trigonometric functions are differentiated using standard trigonometric derivative rules.
Example 3:
Find the first partial derivatives of h(x, y) = cos(xy) + sin(x)
.
-
∂h/∂x: -y sin(xy) + cos(x) (using the chain rule)
-
∂h/∂y: -x sin(xy)
3. Exponential and Logarithmic Functions
Similar to trigonometric functions, we use the standard rules for exponential and logarithmic functions.
Example 4:
Find the first partial derivatives of k(x, y) = ln(x² + y²) + e<sup>xy</sup>
.
-
∂k/∂x: (2x) / (x² + y²) + ye<sup>xy</sup> (using the chain rule)
-
∂k/∂y: (2y) / (x² + y²) + xe<sup>xy</sup> (using the chain rule)
4. Implicit Differentiation
When a function is defined implicitly, we use implicit differentiation. This involves differentiating both sides of the equation with respect to the variable of interest and then solving for the partial derivative.
Example 5:
Find ∂z/∂x and ∂z/∂y for the implicitly defined function x² + y² + z² = 1.
-
∂z/∂x: Differentiate with respect to x: 2x + 2z(∂z/∂x) = 0. Solving for ∂z/∂x gives: ∂z/∂x = -x/z.
-
∂z/∂y: Differentiate with respect to y: 2y + 2z(∂z/∂y) = 0. Solving for ∂z/∂y gives: ∂z/∂y = -y/z.
5. Chain Rule for Partial Derivatives
The chain rule extends to partial derivatives when dealing with composite functions. If z = f(x, y)
and x = g(u, v)
and y = h(u, v)
, then:
- ∂z/∂u = (∂z/∂x)(∂x/∂u) + (∂z/∂y)(∂y/∂u)
- ∂z/∂v = (∂z/∂x)(∂x/∂v) + (∂z/∂y)(∂y/∂v)
Example 6:
Let z = x² + y², x = u cos(v), and y = u sin(v). Find ∂z/∂u and ∂z/∂v.
- ∂z/∂x = 2x
- ∂z/∂y = 2y
- ∂x/∂u = cos(v)
- ∂x/∂v = -u sin(v)
- ∂y/∂u = sin(v)
- ∂y/∂v = u cos(v)
Therefore:
- ∂z/∂u = 2x cos(v) + 2y sin(v) = 2u cos²(v) + 2u sin²(v) = 2u
- ∂z/∂v = 2x(-u sin(v)) + 2y(u cos(v)) = -2u²cos(v)sin(v) + 2u²sin(v)cos(v) = 0
6. Higher-Order Partial Derivatives
After finding first-order partial derivatives, we can proceed to higher-order partial derivatives. For a function f(x, y), we have:
- Second-order partial derivatives: ∂²f/∂x², ∂²f/∂y², ∂²f/∂x∂y (mixed partial derivative), ∂²f/∂y∂x (mixed partial derivative).
For many continuous and differentiable functions, the mixed partial derivatives are equal (Clairaut's theorem): ∂²f/∂x∂y = ∂²f/∂y∂x.
Example 7:
Find the second-order partial derivatives of f(x, y) = x³y² + sin(x)
- ∂f/∂x = 3x²y² + cos(x)
- ∂f/∂y = 2x³y
- ∂²f/∂x² = 6xy² - sin(x)
- ∂²f/∂y² = 2x³
- ∂²f/∂x∂y = 6x²y
- ∂²f/∂y∂x = 6x²y
Applications of Partial Derivatives
Partial derivatives are essential tools in various fields:
- Physics: Calculating rates of change in multivariable systems (e.g., temperature gradients, fluid flow).
- Engineering: Optimizing designs, analyzing stress and strain in materials.
- Economics: Modeling economic systems, analyzing marginal productivity.
- Machine Learning: Gradient descent optimization algorithms rely heavily on partial derivatives to find minima or maxima of cost functions.
- Computer Graphics: Surface normals and shading calculations utilize partial derivatives.
Conclusion
Finding first partial derivatives is a fundamental skill in multivariable calculus. This comprehensive guide has explored various techniques for calculating partial derivatives, including those involving polynomials, trigonometric, exponential, and logarithmic functions. We've also covered implicit differentiation and the application of the chain rule for composite functions. Understanding and mastering these techniques are crucial for tackling more advanced topics in calculus and its numerous applications across diverse fields. Remember to practice consistently to solidify your understanding and build proficiency in these important calculations. The more examples you work through, the more comfortable you will become with different types of functions and techniques.
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