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In Exercises \(1-22,\) find \(\partial f / \partial x\) and \(\partial f / \partial y\) $$f(x, y)=e^{(x+y+1)}$$

Short Answer

Expert verified
Both partial derivatives are \( e^{(x+y+1)} \).

Step by step solution

01

Identify the Function

The function given is \( f(x, y) = e^{(x+y+1)} \). We need to find the partial derivatives \( \partial f / \partial x \) and \( \partial f / \partial y \).
02

Compute the Partial Derivative with respect to x

To find \( \partial f / \partial x \), we treat \( y \) as a constant and differentiate \( f(x, y) = e^{(x+y+1)} \) with respect to \( x \). Since the derivative of \( e^u \) with respect to \( u \) is \( e^u \), and \( u = x+y+1 \), we have:\[ \frac{\partial f}{\partial x} = e^{(x+y+1)} \cdot 1 = e^{(x+y+1)} \].
03

Compute the Partial Derivative with respect to y

Similarly, for \( \partial f / \partial y \), we treat \( x \) as a constant and differentiate \( f(x, y) = e^{(x+y+1)} \) with respect to \( y \). Using the same rule as before, we have:\[ \frac{\partial f}{\partial y} = e^{(x+y+1)} \cdot 1 = e^{(x+y+1)} \].
04

Conclusion

Both partial derivatives result in the same expression, thus:\( \frac{\partial f}{\partial x} = e^{(x+y+1)} \) and \( \frac{\partial f}{\partial y} = e^{(x+y+1)} \).

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Key Concepts

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

Partial Derivative with respect to x
When dealing with functions of multiple variables, such as \( f(x, y) = e^{(x+y+1)} \), we often need to explore how these functions change with respect to one particular variable while treating others as constants. This concept is central to understanding partial derivatives. In mathematics, taking the partial derivative of a function with respect to \( x \) means you differentiate as usual but treat all other variables as constants. This helps in understanding how the function changes along the \( x \)-axis.

For example, when finding \( \frac{\partial f}{\partial x} \) for \( f(x, y) = e^{(x+y+1)} \), the approach is straightforward:
  • Identify the variable of interest, which is \( x \).
  • Treat \( y \) as a constant throughout the differentiation process.
  • Apply the derivative rule for exponential functions.
Thus, the partial derivative is found exactly as in the solution, resulting in \( e^{(x+y+1)} \), showing how the function changes when just \( x \) is varied.
Partial Derivative with respect to y
The partial derivative with respect to \( y \) follows a similar process as with \( x \), but this time, \( x \) is treated as a constant. This allows us to understand how the function behaves when only the variable \( y \) is changed. Such derivatives are crucial when working with multi-variable functions, significantly simplifying complex problems by isolating the impact of one variable.

For the function \( f(x, y) = e^{(x+y+1)} \):
  • We focus on \( y \) and consider \( x \) as a constant.
  • Then, differentiate using the exponential rule as for single-variable functions.
Here, this results in the equation for \( \frac{\partial f}{\partial y} \) becoming \( e^{(x+y+1)} \) as well. It's insightful to note how both partial derivatives being equal highlights the symmetry in the function's structure concerning \( x \) and \( y \).
Exponential Functions
Exponential functions, like \( e^{(x+y+1)} \), are a significant type of mathematical function frequently encountered across various fields such as science, engineering, and finance. An exponential function is one where the variable appears in the exponent, leading to unique properties and behaviors.

Key characteristics include:
  • The base, in this case \( e \) (Euler's number), is a constant approximately equal to 2.71828.
  • They grow very quickly and exhibit continuous and smooth curves across their domain.

When differentiating exponential functions, the exponential term itself does not change due to its derivative being itself, meaning \( \frac{d}{du}(e^u) = e^u \). This is particularly handy when taking partial derivatives, since the exponential component remains the same, simplifying calculations and allowing focus on the linear component in the exponent, such as \( x+y+1 \) in our function. Understanding these basic properties aids in manipulating and effectively applying exponential functions in various contexts.

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Most popular questions from this chapter

In Exercises \(49-52\) , find an equation for and sketch the graph of the level curve of the function \(f(x, y)\) that passes through the given point. $$ f(x, y)=16-x^{2}-y^{2}, \quad(2 \sqrt{2}, \sqrt{2}) $$

You will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level curves plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. d. Calculate the function's second partial derivatives and find the discriminant \(f_{x x} f_{y y}-f_{x y}^{2}\) . e. Using the max-min tests, classify the critical points found in part (c). Are your findings consistent with your discussion in part (c)? $$f(x, y)=x^{3}-3 x y^{2}+y^{2},-2 \leq x \leq 2,-2 \leq y \leq 2$$

In Exercises \(53-60,\) sketch a typical level surface for the function. $$ f(x, y, z)=y^{2}+z^{2} $$

Parametrized Surfaces Just as you describe curves in the plane parametrically with a pair of equations \(x=f(t), y=g(t)\) defined on some parameter interval \(I,\) you can sometimes describe surfaces in space with a triple of equations \(x=f(u, v), y=g(u, v), z=h(u, v)\) defined on some parameter rectangle \(a \leq u \leq b, c \leq v \leq d .\) Many computer algebra systems permit you to plot such surfaces in parametric mode. (Parametrized surfaces are discussed in detail in Section 16.5.) Use a CAS to plot the surfaces in Exercises \(77-80 .\) Also plot several level curves in the \(x y\) -plane. $$ \begin{array}{l}{x=u \cos v, \quad y=u \sin v, \quad z=u, \quad 0 \leq u \leq 2} \\ {0 \leq v \leq 2 \pi}\end{array} $$

Two dependent variables Find \(\partial x / \partial u\) and \(\partial y / \partial u\) if the equations \(u=x^{2}-y^{2}\) and \(v=x^{2}-y\) define \(x\) and \(y\) as functions of the independent variables \(u\) and \(v,\) and the partial derivatives exist. (See the hint in Exercise \(69 . )\) Then let \(s=x^{2}+y^{2}\) and find \(\partial s / \partial u .\)

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