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Chapter 10: Problem 11

Find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. \(f(x, y)=e^{x} \sin (x y)\)

Short Answer

Expert verified
\(\frac{\partial f}{\partial x} = e^x \sin(xy) + e^x y \cos(xy)\) and \(\frac{\partial f}{\partial y} = e^x x \cos(xy)\).

Step by step solution

01

Understand the Problem

We need to find the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) for the function \(f(x, y) = e^x \sin(xy)\). This involves treating each variable independently while taking the derivative.
02

Find \(\frac{\partial f}{\partial x}\)

To find the partial derivative with respect to \(x\), differentiate \(f(x, y) = e^x \sin(xy)\) with respect to \(x\) while treating \(y\) as a constant. Use the product rule: \((uv)' = u'v + uv'\), where \(u = e^x\) and \(v = \sin(xy)\). So, \(u' = e^x\) and \(v' = y\cos(xy)\).Apply the product rule:\[\frac{\partial f}{\partial x} = e^x \cdot \sin(xy) + e^x y \cos(xy)\]
03

Find \(\frac{\partial f}{\partial y}\)

To find the partial derivative with respect to \(y\), differentiate \(f(x, y) = e^x \sin(xy)\) with respect to \(y\) while treating \(x\) as a constant. The derivative of \(\sin(xy)\) with respect to \(y\) is \(x \cos(xy)\).Hence:\[\frac{\partial f}{\partial y} = e^x \cdot x \cos(xy)\]
04

Write the Results

Summarize the derivatives:The partial derivative of \(f\) with respect to \(x\) is:\[\frac{\partial f}{\partial x} = e^x \sin(xy) + e^x y \cos(xy)\]The partial derivative of \(f\) with respect to \(y\) is:\[\frac{\partial f}{\partial y} = e^x x \cos(xy)\]

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Key Concepts

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

Partial Derivatives
When working with functions of multiple variables, such as the given function \( f(x, y) = e^x \sin(xy) \), we often need to explore how changes in one variable affect the function while keeping the other variable constant. This exploration leads us to partial derivatives.

A partial derivative of a function is essentially the derivative of the function with respect to one of its variables, treating all other variables as constants. For example:
  • The partial derivative \( \frac{\partial f}{\partial x} \) provides the rate of change of \( f \) when \( x \) changes and \( y \) remains constant.
  • The partial derivative \( \frac{\partial f}{\partial y} \) does the same but for changes in \( y \) while \( x \) stays constant.

By understanding partial derivatives, we can better analyze and model real-world problems involving more than one variable, such as in engineering or physics.
Product Rule
In calculus, the product rule is a key method for finding the derivative of a product of two functions. It states that if we have two differentiable functions \( u(x) \) and \( v(x) \), the derivative of their product \( u(x)v(x) \) with respect to \( x \) is given by \[ \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \]

When we apply the product rule in the context of partial derivatives, we treat one variable as the primary variable and others as constants. In our exercise, for \( f(x, y) = e^x \sin(xy) \), we identified the parts of the function as:
  • \( u = e^x \) and \( v = \sin(xy) \)
  • \( u' = e^x \) and \( v' = y\cos(xy) \)
Applying the product rule helped us find that:
  • \( \frac{\partial f}{\partial x} = e^x \sin(xy) + e^x y \cos(xy) \)
This tool is particularly useful when dealing with complex expressions involving products, making the process of finding derivatives manageable.
Trigonometric Functions
Trigonometric functions, like \( \sin \) and \( \cos \), often appear in calculus problems involving oscillatory or periodic behavior. In the given function, \( f(x, y) = e^x \sin(xy) \), \( \sin(xy) \) is the trigonometric component.

Trigonometric functions have well-known derivatives:
  • The derivative of \( \sin(x) \) is \( \cos(x) \)
  • The derivative of \( \cos(x) \) is \(-\sin(x) \)
In partial derivatives, you apply these rules by differentiating with respect to one variable while holding others constant. For example, in \( \frac{\partial f}{\partial y} \), the derivative of \( \sin(xy) \) with respect to \( y \) translates to :
  • \( x \cos(xy) \).
These rules simplify the process of working with trigonometric expressions, helping to unravel their role in more complex functions. Recognizing their patterns quickly becomes a vital skill in calculus.

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Most popular questions from this chapter

Give a geometric interpretation of the set $$A=\left\\{(x, y) \in \mathbf{R}^{2}: \sqrt{x^{2}+6 x+y^{2}-2 y+10}<2\right\\}$$

(a) Write $$h(x, y)=\cos (y-x)$$ as a composition of two functions. (b) For which values of \((x, y)\) is \(h(x, y)\) continuous?

Show that $$ c(x, t)=\frac{1}{\sqrt{8 \pi t}} \exp \left[-\frac{x^{2}}{8 t}\right] $$ solves $$ \frac{\partial c(x, t)}{\partial t}=2 \frac{\partial^{2} c(x, t)}{\partial x^{2}} $$

Morphogenesis Most embryos start out as lumps of cells. Cells in these lumps are initially undifferentiated-that is, they start out all in the same state. Over time cells then commit to different functions, e.g., to becoming legs, eyes, and so on. To do this chemicals called morphogens are distributed unequally through the embryo, allowing each cell to tell where in the embryo it is located. How are unequal distributions of morphogens achieved? One model for how morphogens can be distributed through the embryo is that morphogens are continuously produced at one end (also called pole) of the embryo. From there they diffuse through the embryo. As the morphogens diffuse, they are constantly broken down by the cells in the embryo. First let's ignore the process of morphogen degradation, and focus only on diffusion. We will assume that the pole at which the morphogen is produced is located at \(x=0 ;\) and for simplicity's sake the cell occupies the interval \(x \geq 0\). Then our partial differential equation model for the distribution of morphogen becomes: $$ \frac{\partial c}{\partial t}=D \frac{\partial^{2} c}{\partial x^{2}} \quad \text { for } \quad x>0 $$ with $$ -D \frac{\partial c}{\partial x}(0, t)=Q \quad \text { and } \quad c(x, t) \rightarrow 0 \text { as } x \rightarrow \infty $$ where \(Q\) is the rate of morphogen production. (a) Let's try to find a steady state distribution of morphogen. That is, we will assume that over time the morphogen concentration reaches some state that does not change with time, i.e., the concentration is given by a function \(C(x) .\) Then \(C(x)\) will satisfy the partial differential equation if and only if: $$ \begin{aligned} 0 &=D \frac{d^{2} C}{d x^{2}} \quad \text { for } \quad x>0 \\ -D C^{\prime}(0) &=Q \quad \text { and } \quad C(x) \rightarrow 0 \text { as } x \rightarrow \infty \end{aligned} $$ Show that there is no function \(C(x)\) that satisfies this differential equation. [Hint: Start by integrating once \((10.48)\) to find \(d C / d x\) and then again to find \(\mathcal{C}(x)\), then try to impose the constraints at \(x=0\), and as \(x \rightarrow \infty\) on your solution.] (b) Now let's incorporate morphogen degradation into our model. We will assume that the breakdown of morphogen has first order kinetics (see Section \(5.9\) for a discussion of the different kinds of kinetics that chemical reactions may have). This means that in one unit of time a fraction \(r\) of the morphogen contained in each region of the embryo is degraded. Then our partial differential equation must be altered to: $$ \frac{\partial c}{\partial t}=D \frac{\partial^{2} c}{\partial x^{2}}-r c \quad \text { for } \quad x>0 $$ with $$ -D \frac{\partial c}{\partial x}(0, t)=Q \quad \text { and } \quad c(x, t) \rightarrow 0 \text { as } x \rightarrow \infty $$ Show that this partial differential equation does have a steady state solution of the form: $$ C(x)=Q \sqrt{\frac{1}{D r}} \exp \left(-\sqrt{\frac{r}{D}} x\right) $$ That is, check that this function \(C(x)\) satisfies both the steady state form of \((10.49)\) as well as the constraints at \(x=0\) and as \(x \rightarrow \infty\).

Find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. \(f(x, y)=x^{2} y+x y^{2}\)
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