Problem 2
Question
Find the first partial derivatives with respect to \(x\) and with respect to \(y\). $$ z=x^{2}-2 y $$
Step-by-Step Solution
Verified Answer
The first partial derivative of \(z=x^{2}-2y\) with respect to \(x\) is \(\frac{\partial z}{\partial x} = 2x\) and with respect to \(y\) is \(\frac{\partial z}{\partial y} = -2\).
1Step 1: Find the partial derivative with respect to \(x\)
To find the partial derivative of the given function with respect to \(x\), differentiate the function treating all other variables (in this case just \(y\)) as constants. Applying the power rule for differentiation, we get \(\frac{\partial z}{\partial x} = 2x\).
2Step 2: Find the partial derivative with respect to \(y\)
Similarly, to find the partial derivative with respect to \(y\), we differentiate the function treating all other variables (in this case \(x\)) as constants. Because \(y\) appears in the function as a linear term, its derivative is constant and we get \(\frac{\partial z}{\partial y} = -2\)
Key Concepts
Power RuleDifferentiation with Respect to xDifferentiation with Respect to y
Power Rule
The power rule is a basic tool for differentiating functions of the form \(x^n\), where \(n\) is a real number. It states that the derivative of \(x^n\) with respect to \(x\) is \(nx^{n-1}\). This rule greatly simplifies many differentiation problems by giving a straightforward method to follow.
In the context of our exercise, when applying the power rule to \(x^2\), \(n\) is 2. Thus, the derivative becomes \(2x^{2-1} = 2x\). This rule is crucial when finding partial derivatives, as it allows us to efficiently calculate the rate of change of functions with powers of variables.
In the context of our exercise, when applying the power rule to \(x^2\), \(n\) is 2. Thus, the derivative becomes \(2x^{2-1} = 2x\). This rule is crucial when finding partial derivatives, as it allows us to efficiently calculate the rate of change of functions with powers of variables.
Differentiation with Respect to x
When you differentiate a function with respect to \(x\), you consider all other variables as constants. This means you focus on how the function changes with every small change in \(x\), keeping everything else constant.
In our exercise, the function \(z = x^2 - 2y\) requires finding the partial derivative with respect to \(x\). Applying the power rule to \(x^2\), we treat \(-2y\) as a constant since it doesn't change with \(x\). Therefore, the partial derivative with respect to \(x\) is \(\frac{\partial z}{\partial x} = 2x\). Practicing this helps in understanding the influence of each variable separately in multi-variable functions.
In our exercise, the function \(z = x^2 - 2y\) requires finding the partial derivative with respect to \(x\). Applying the power rule to \(x^2\), we treat \(-2y\) as a constant since it doesn't change with \(x\). Therefore, the partial derivative with respect to \(x\) is \(\frac{\partial z}{\partial x} = 2x\). Practicing this helps in understanding the influence of each variable separately in multi-variable functions.
Differentiation with Respect to y
Differentiating with respect to \(y\) means viewing all other terms as constants. You observe how the function changes with slight changes in \(y\), while keeping \(x\) fixed.
For the function \(z = x^2 - 2y\), to find the partial derivative with respect to \(y\), we differentiate \(-2y\) and treat \(x^2\) as a constant. Since \(y\) appears as a linear term, its derivative is simply \(-2\). Thus, the partial derivative with respect to \(y\) is \(\frac{\partial z}{\partial y} = -2\). This process highlights how each variable individually contributes to the overall change in functions.
For the function \(z = x^2 - 2y\), to find the partial derivative with respect to \(y\), we differentiate \(-2y\) and treat \(x^2\) as a constant. Since \(y\) appears as a linear term, its derivative is simply \(-2\). Thus, the partial derivative with respect to \(y\) is \(\frac{\partial z}{\partial y} = -2\). This process highlights how each variable individually contributes to the overall change in functions.
Other exercises in this chapter
Problem 2
Use Lagrange multipliers to find the given extremum. In each case, assume that \(x\) and \(y\) are positive. $$ \text { Maximize } f(x, y)=x y \quad 2 x+y=4 $$
View solution Problem 2
Find any critical points and relative extrema of the function. $$ f(x, y)=x^{2}+y^{2}+2 x-6 y+6 $$
View solution Problem 2
Find the function values. \(f(x, y)=4-x^{2}-4 y^{2}\) (a) \(f(0,0)\) (b) \(f(0,1)\) (c) \(f(2,3)\) (d) \(f(1, y)\) (e) \(f(x, 0)\) (f) \(f(t, 1)\)
View solution Problem 2
Find the intercepts and sketch the graph of the plane. $$ 3 x+6 y+2 z=6 $$
View solution