Problem 10 Find the first partial derivativ... [FREE SOLUTION]

Chapter 16: Problem 10

Find the first partial derivatives of $f$. $$ f(x, y)=\sqrt{4 x^{2}-y^{2}} \sec x $$

Short Answer

Expert verified

The partial derivatives are $ \frac{\partial f}{\partial x} = \frac{4x}{\sqrt{4x^2 - y^2}} \sec x + (\sqrt{4x^2 - y^2}) \sec x \tan x $ and $ \frac{\partial f}{\partial y} = \frac{-y}{\sqrt{4x^2 - y^2}} \sec x $.

Step by step solution

Identify the Components of the Function

The given function is $ f(x, y) = \sqrt{4x^2 - y^2} \sec x $. This function is comprised of two parts: $ \sqrt{4x^2 - y^2} $, which is a function of both $ x $ and $ y $, and $ \sec x $, which is a function of $ x $. Our goal is to find the partial derivatives with respect to $ x $ and $ y $.

Differentiate with Respect to $ x $

To find $ \frac{\partial f}{\partial x} $, treat $ y $ as a constant and apply the product rule: $(uv)' = u'v + uv'$, where $ u = \sqrt{4x^2 - y^2} $ and $ v = \sec x $. First, differentiate $ u = \sqrt{4x^2 - y^2} $ with respect to $ x $:\[\frac{\partial u}{\partial x} = \frac{1}{2\sqrt{4x^2 - y^2}} \cdot 8x = \frac{4x}{\sqrt{4x^2 - y^2}}\]Next, differentiate $ v = \sec x $ with respect to $ x $:\[\frac{d}{dx}(\sec x) = \sec x \tan x\]Now, apply the product rule:\[\frac{\partial f}{\partial x} = \left(\frac{4x}{\sqrt{4x^2 - y^2}}\right) \sec x + \sqrt{4x^2 - y^2} (\sec x \tan x)\]

Differentiate with Respect to $ y $

To find $ \frac{\partial f}{\partial y} $, treat $ x $ as a constant. Only the $ \sqrt{4x^2 - y^2} $ part depends on $ y $.Differentiate $ \sqrt{4x^2 - y^2} $ with respect to $ y $:\[\frac{\partial}{\partial y} \left(\sqrt{4x^2 - y^2}\right) = \frac{1}{2\sqrt{4x^2 - y^2}} \cdot (-2y) = \frac{-y}{\sqrt{4x^2 - y^2}}\]The $ \sec x $ remains as it is independent of $ y $:\[\frac{\partial f}{\partial y} = \frac{-y}{\sqrt{4x^2 - y^2}} \sec x\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule in Calculus

When you have a function that is the product of two functions, the product rule becomes handy in calculus. It helps us differentiate functions that are multiplied together. Imagine you have a scenario where one part of the function depends on a variable, while the other part might not, or varies differently.

The basic formula of the product rule is: $(uv)' = u'v + uv'$.
This means you first differentiate the first function while keeping the second one constant and then do the opposite, adding the results together.

In our exercise, we applied the product rule to the components $ u = \sqrt{4x^2 - y^2} $ and $ v = \sec x $. By treating $ y $ as a constant when differentiating with respect to $ x $, it allows us to cleanly apply this rule. This approach helps us find partial derivatives effectively when dealing with multivariable functions.

Understanding Trigonometric Functions

Trigonometric functions are functions of angles and are fundamental in the study of calculus and they often appear in various equations and formulas. In our exercise, $ \sec x $ is a trigonometric function, which is equivalent to $ 1/\cos x $.

The derivative of $ \sec x $ is $ \sec x \tan x $, a key identity in differentiation.
These derivatives tell us how the secant and tangent functions change as $ x $ changes.

When calculating partial derivatives involving trigonometric functions, it's crucial to recognize these identities. Knowing these simple derivatives makes the process much easier and avoids errors in computation. It also illustrates how angles and rates of change are interconnected in calculus.

Basics of Calculus and Partial Derivatives

Calculus is the branch of mathematics that focuses on change and motion through derivatives and integrals. Partial derivatives are particularly useful when we deal with functions of several variables as they measure how a function changes as one particular variable changes, while others are held constant.

When dealing with partial derivatives, treat all other variables as constants.
They help to understand how multi-variable functions behave under changes in specific variables.

In the exercise, we found the first partial derivatives $ \frac{\partial f}{\partial x} $ and $ \frac{\partial f}{\partial y} $ by focusing on how the function varies with changes in $ x $ and $ y $, respectively. Understanding partial derivatives is crucial for solving real-world problems involving functions with multiple variables.

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91影视

Find the first partial derivatives of \(f\). $$ f(x, y)=\sqrt{4 x^{2}-y^{2}} \sec x $$

Short Answer

Step by step solution

Identify the Components of the Function

Differentiate with Respect to \( x \)

Differentiate with Respect to \( y \)

Key Concepts

Product Rule in Calculus

Understanding Trigonometric Functions

Basics of Calculus and Partial Derivatives

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