Problem 39
Question
Find the indicated partial derivatives. \(f(x, y)=x^{2} y^{2}+2 x y^{2} ; \frac{a^{2} f}{\partial x^{2}}\)
Step-by-Step Solution
Verified Answer
The second partial derivative \(\frac{\partial^2 f}{\partial x^2}\) is \(2y^2\).
1Step 1: Understand the Function
We are given the function \( f(x, y) = x^2 y^2 + 2xy^2 \). The function is defined with two variables: \(x\) and \(y\).
2Step 2: Differentiate with Respect to x (First Derivative)
To find the partial derivative of \(f\) with respect to \(x\), consider \(y\) as a constant. Differentiate each term separately: - The derivative of \(x^2 y^2\) with respect to \(x\) is \(2xy^2\).- The derivative of \(2xy^2\) with respect to \(x\) is \(2y^2\).The first derivative \(\frac{\partial f}{\partial x}\) is: \[\frac{\partial f}{\partial x} = 2xy^2 + 2y^2\].
3Step 3: Differentiate Again with Respect to x (Second Derivative)
Now take the derivative of \(\frac{\partial f}{\partial x} = 2xy^2 + 2y^2\) with respect to \(x\), again treating \(y\) as a constant:- The derivative of \(2xy^2\) with respect to \(x\) is \(2y^2\).- The derivative of \(2y^2\) with respect to \(x\) is \(0\), since it is a constant in terms of \(x\).The second derivative, \(\frac{\partial^2 f}{\partial x^2}\), is:\[\frac{\partial^2 f}{\partial x^2} = 2y^2\].
Key Concepts
Multivariable CalculusDifferentiationSecond Derivative
Multivariable Calculus
Multivariable Calculus is an extension of traditional calculus, where we study functions with more than one variable. In this context, functions can be defined in terms of more than one independent variable, like in our example function:
\[ f(x, y) = x^2 y^2 + 2xy^2 \]
Here, both \(x\) and \(y\) are variables that can change independently. Multivariable calculus focuses on understanding how changes in these variables affect the function's overall behavior.
\[ f(x, y) = x^2 y^2 + 2xy^2 \]
Here, both \(x\) and \(y\) are variables that can change independently. Multivariable calculus focuses on understanding how changes in these variables affect the function's overall behavior.
- Partial Derivatives: These help us measure how a function changes as one specific variable changes, while keeping others constant.
- Applications: Useful in fields like engineering and physics to model complex systems where multiple factors interact.
Differentiation
Differentiation is a key concept in calculus that deals with finding the rate of change of a function. When applied to multivariable calculus, we use partial differentiation, where we take the derivative with respect to one variable while treating others as constants.
For the given function \(f(x, y)\), we analyzed how \(f\) changes concerning \(x\), leading to the first partial derivative \(\frac{\partial f}{\partial x}\). This step involves straightforward computation of derivatives:
Partial derivatives play a crucial role in studying surfaces, gradients, and optimization problems in multiple dimensions.
For the given function \(f(x, y)\), we analyzed how \(f\) changes concerning \(x\), leading to the first partial derivative \(\frac{\partial f}{\partial x}\). This step involves straightforward computation of derivatives:
- First Derivative: Differentiating \(x^2 y^2\) gives \(2xy^2\), treating \(y^2\) as a constant.
- First Derivative: Differentiating \(2xy^2\) yields \(2y^2\), again considering \(y^2\) constant.
Partial derivatives play a crucial role in studying surfaces, gradients, and optimization problems in multiple dimensions.
Second Derivative
The second derivative is a further step in understanding the behavior of a function, providing insights into its concavity and rate of variation changes. For functions involving multiple variables, the second partial derivative can be a bit more nuanced.
In our task, after finding the first derivative, we proceeded to find the second derivative \(\frac{\partial^2 f}{\partial x^2}\), which tells us about the curvature of the function relating \(x\).
In our task, after finding the first derivative, we proceeded to find the second derivative \(\frac{\partial^2 f}{\partial x^2}\), which tells us about the curvature of the function relating \(x\).
- The Second Derivative of \(2xy^2\): Differentiating once more with respect to \(x\) gives \(2y^2\), highlighting constant slope changes as \(y^2\) remains unaltered.
- The Constant Term \(2y^2\): Its derivative remains \(0\) since it does not vary with \(x\).
Other exercises in this chapter
Problem 38
Find a linear approximation to each func\mathrm{tion } \(f(x, y)\) at the indicated point. \(\mathbf{f}(x, y)=\left[\begin{array}{c}3 x-y^{2} \\ 4 y\end{array}\
View solution Problem 38
Use Lagrange multipliers to find the maxima and minima of the functions under the given constraints. $$f(x, y)=x y ; x+y=4$$
View solution Problem 39
Determine the equation of the level curves \(f(x, y)=c\) and sketch the level curves for the specified values of \(c\). \(f(x, y)=x+y ; c=0,-1,1\)
View solution Problem 39
Use Lagrange multipliers to find the maxima and minima of the functions under the given constraints. $$ f(x, y)=x y ; x^{2}+4 y^{2}=1 $$
View solution