Problem 14
Question
In Problems 1-16, find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. $$ f(x, y)=\ln \left(3 x^{2}-x y\right) $$
Step-by-Step Solution
Verified Answer
\( \frac{\partial f}{\partial x} = \frac{6x - y}{3x^2 - xy} \) and \( \frac{\partial f}{\partial y} = \frac{-x}{3x^2 - xy} \).
1Step 1: Identify the Function
The given function is \( f(x, y) = \ln(3x^2 - xy) \). We need to find the partial derivatives \( \partial f / \partial x \) and \( \partial f / \partial y \).
2Step 2: Find \( \partial f / \partial x \)
To find \( \frac{\partial f}{\partial x} \), differentiate the function with respect to \( x \) while treating \( y \) as a constant.Using the chain rule, the derivative of \( \ln(u) \) is \( \frac{1}{u} \frac{\partial u}{\partial x} \). Here, \( u = 3x^2 - xy \).Compute \( \frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(3x^2 - xy) = 6x - y \).Thus,\[\frac{\partial f}{\partial x} = \frac{1}{3x^2 - xy}(6x - y)\].
3Step 3: Find \( \partial f / \partial y \)
To find \( \frac{\partial f}{\partial y} \), differentiate the function with respect to \( y \) while treating \( x \) as a constant.Using the chain rule, the derivative of \( \ln(u) \) is \( \frac{1}{u} \frac{\partial u}{\partial y} \). With \( u = 3x^2 - xy \), compute \( \frac{\partial u}{\partial y} = \frac{\partial}{\partial y}(3x^2 - xy) = -x \).Therefore,\[\frac{\partial f}{\partial y} = \frac{1}{3x^2 - xy}(-x)\].
4Step 4: Write Final Answers
The partial derivatives are:\[\frac{\partial f}{\partial x} = \frac{6x - y}{3x^2 - xy}\]\[\frac{\partial f}{\partial y} = \frac{-x}{3x^2 - xy}\]
Key Concepts
Chain Rule and Its ApplicationUnderstanding Multivariable CalculusExploring the Natural Logarithm
Chain Rule and Its Application
The Chain Rule is an essential tool in calculus, especially when dealing with composite functions. It helps us differentiate functions of a function, like when you have an outer function, such as a natural logarithm, and an inner function, like a polynomial.
In this case, our function is \( f(x, y) = \ln(3x^2 - xy) \). Here, the outer function is the natural logarithm \( \ln(u) \), and the inner function \( u \) is \( 3x^2 - xy \). To differentiate the outer function, we apply the chain rule: the derivative of \( \ln(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \).
- First, differentiate the outer function with respect to \( u \), which gives us \( \frac{1}{u} \).- Then, differentiate the inner function \( u \) with respect to \( x \) or \( y \).This rule allows us to compute partial derivatives of multivariable functions with ease, primarily when the functions involve nested expressions. The chain rule ensures we are consistent in how we handle these interdependent variables.
In this case, our function is \( f(x, y) = \ln(3x^2 - xy) \). Here, the outer function is the natural logarithm \( \ln(u) \), and the inner function \( u \) is \( 3x^2 - xy \). To differentiate the outer function, we apply the chain rule: the derivative of \( \ln(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \).
- First, differentiate the outer function with respect to \( u \), which gives us \( \frac{1}{u} \).- Then, differentiate the inner function \( u \) with respect to \( x \) or \( y \).This rule allows us to compute partial derivatives of multivariable functions with ease, primarily when the functions involve nested expressions. The chain rule ensures we are consistent in how we handle these interdependent variables.
Understanding Multivariable Calculus
Multivariable Calculus extends the concepts of calculus to functions of several variables. This field encompasses techniques like partial differentiation, which we applied in the exercise.
Partial derivatives are similar to regular derivatives, but they measure how a function changes as just one of the inputs changes, keeping all others constant. This makes them pivotal in dealing with multivariable functions, offering insights on how changes in one variable affect the overall function.
- In our exercise, we found the partial derivatives \( \partial f / \partial x \) and \( \partial f / \partial y \).- By treating all other variables constant, partial derivatives provide a window into the function's behavior concerning each variable independently.In multivariable calculus, these derivatives are useful in optimization problems and predicting how quantities interact in complex systems. Mastery of these concepts equips you to tackle real-world problems where multiple variables influence outcomes simultaneously.
Partial derivatives are similar to regular derivatives, but they measure how a function changes as just one of the inputs changes, keeping all others constant. This makes them pivotal in dealing with multivariable functions, offering insights on how changes in one variable affect the overall function.
- In our exercise, we found the partial derivatives \( \partial f / \partial x \) and \( \partial f / \partial y \).- By treating all other variables constant, partial derivatives provide a window into the function's behavior concerning each variable independently.In multivariable calculus, these derivatives are useful in optimization problems and predicting how quantities interact in complex systems. Mastery of these concepts equips you to tackle real-world problems where multiple variables influence outcomes simultaneously.
Exploring the Natural Logarithm
The natural logarithm, denoted as \( \ln(x) \), is a logarithm with base \( e \), where \( e \approx 2.71828 \). It plays a considerable role in calculus due to its properties that simplify differentiation and integration processes.
One of the distinguished properties of the natural logarithm is its derivative. When differentiating \( \ln(u) \) with respect to another variable, we employ the chain rule, yielding the derivative \( \frac{1}{u} \cdot \frac{du}{dx} \). This simplifies calculations when \( u \) itself is a more complex expression.
- The natural logarithm transforms multiplicative relationships into additive ones.- It helps in linearizing exponential growth, making complex mathematical patterns more manageable.In functions involving the natural logarithm, understanding this logarithmic behavior facilitates tackling equations that emerge in diverse scientific fields, from finance to physics. Comprehending how natural logarithms integrate into calculus enriches your mathematical toolkit, bringing clarity to intricate exponential relationships.
One of the distinguished properties of the natural logarithm is its derivative. When differentiating \( \ln(u) \) with respect to another variable, we employ the chain rule, yielding the derivative \( \frac{1}{u} \cdot \frac{du}{dx} \). This simplifies calculations when \( u \) itself is a more complex expression.
- The natural logarithm transforms multiplicative relationships into additive ones.- It helps in linearizing exponential growth, making complex mathematical patterns more manageable.In functions involving the natural logarithm, understanding this logarithmic behavior facilitates tackling equations that emerge in diverse scientific fields, from finance to physics. Comprehending how natural logarithms integrate into calculus enriches your mathematical toolkit, bringing clarity to intricate exponential relationships.
Other exercises in this chapter
Problem 14
Show that \(f(x, y)\) is differentiable at the indicated point. $$ f(x, y)=e^{x-y} ;(0,0) $$
View solution Problem 14
Find the largest possible domain and the corresponding range of each function. Determine the equation of the level curves \(f(x, y)=c\), together with the possi
View solution Problem 14
The functions are defined on the rectangular domain $$ D=\\{(x, y):-1 \leq x \leq 1,-1 \leq y \leq 1\\} $$ Find the absolute maxima and minima of \(f\) on \(D\)
View solution Problem 14
In Problems 1-14, use the properties of limits to calculate the following limits: $$ \lim _{(x, y) \rightarrow(1,-2)} \frac{2 x^{2}+y}{2 x y+3} $$
View solution