Chapter 9: Problem 17
Find the first partial derivatives of the given function. $$ z=\frac{4 \sqrt{x}}{3 y^{2}+1} $$
Short Answer
Expert verified
The partial derivatives are \( \frac{2}{\sqrt{x}(3y^2 + 1)} \) and \( \frac{-24y\sqrt{x}}{(3y^2 + 1)^2} \).
Step by step solution
01
Identify the function
We are given the function \(z\) which is defined as \( z = \frac{4 \sqrt{x}}{3y^2 + 1} \). Our task is to find the first partial derivatives with respect to \( x \) and \( y \).
02
Find the partial derivative with respect to x
To find \( \frac{\partial z}{\partial x} \), we treat \( y \) as a constant and differentiate \( z \) with respect to \( x \). Using the quotient rule for derivatives, where \( \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x) v(x) - u(x) v'(x)}{(v(x))^2} \), let \( u(x) = 4\sqrt{x} \) and \( v(y) = 3y^2 + 1 \). Thus, \( u'(x) = \frac{2}{\sqrt{x}} \) and \( v(y) \) is a constant when differentiating with respect to \( x \). We get: \[\frac{\partial z}{\partial x} = \frac{\frac{2}{\sqrt{x}}(3y^2 + 1)}{(3y^2 + 1)^2} = \frac{2}{\sqrt{x}(3y^2 + 1)}.\]
03
Find the partial derivative with respect to y
To find \( \frac{\partial z}{\partial y} \), treat \( x \) as a constant and differentiate \( z \) with respect to \( y \) using the quotient rule. In this case, \( u(x) = 4\sqrt{x} \) (which is constant when differentiating with respect to \( y \)) and \( v(y) = 3y^2 + 1 \), so \( v'(y) = 6y \). Applying the quotient rule, we have: \[\frac{\partial z}{\partial y} = \frac{0 \cdot (3y^2 + 1) - 4\sqrt{x} \cdot 6y}{(3y^2 + 1)^2} = \frac{-24y\sqrt{x}}{(3y^2 + 1)^2}.\]
04
Conclusion
The first partial derivatives of the function \( z = \frac{4 \sqrt{x}}{3y^2 + 1} \) are: \[\frac{\partial z}{\partial x} = \frac{2}{\sqrt{x}(3y^2 + 1)}\]and \[\frac{\partial z}{\partial y} = \frac{-24y\sqrt{x}}{(3y^2 + 1)^2}.\]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Rule
The quotient rule is essential when differentiating functions represented as a fraction. This rule states that if you have a function expressed as a quotient such as \( \frac{u(x)}{v(x)} \), where both \( u \) and \( v \) are differentiable functions, the derivative is given by:\[\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x) v(x) - u(x) v'(x)}{(v(x))^2}\]In our exercise, the function \( z = \frac{4 \sqrt{x}}{3y^2 + 1} \) can be separated into these components:
- \( u(x) = 4 \sqrt{x} \)
- \( v(x) = 3y^2 + 1 \)
Function of Two Variables
A function of two variables is a mathematical expression involving two separate variables which can affect the outcome of the function. In our scenario, the function is:\[ z = \frac{4 \sqrt{x}}{3y^2 + 1} \]Here, \( x \) and \( y \) are the two variables. Understanding functions of this kind is crucial because they allow us to calculate rates of change in multiple directions.With two variables:
- \( x \) affects the square root in the numerator.
- \( y \) influences the polynomial in the denominator.
First Order Derivatives
First order derivatives are essential tools used to understand how a function changes as its inputs change. For functions of two variables, this involves partial derivatives:
- \( \frac{\partial z}{\partial x} \): Represents how \( z \) changes as \( x \) changes, keeping \( y \) constant.
- \( \frac{\partial z}{\partial y} \): Shows the change in \( z \) as \( y \) changes, with \( x \) held constant.