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Find the first partial derivatives of the following functions. $$h(w, x, y, z)=\frac{w z}{x y}$$

Short Answer

Expert verified
Question: Find the first partial derivatives of the function $$h(w,x,y,z) = \frac{wz}{xy}$$ with respect to w, x, y, and z. Answer: The first partial derivatives of the given function are: \[\frac{\partial h}{\partial w} = \frac{z}{xy}, \quad \frac{\partial h}{\partial x} = -\frac{wz}{x^2 y}, \quad \frac{\partial h}{\partial y} = -\frac{wxz}{y^2 x}, \quad \frac{\partial h}{\partial z} = \frac{w}{xy}\]

Step by step solution

01

Identify the given function

The given function is: $$h(w, x, y, z) = \frac{wz}{xy}$$
02

Find the partial derivative with respect to w

We will now find the partial derivative of h with respect to w, while treating x, y, and z as constants. This is denoted as: $$\frac{\partial h}{\partial w}$$ Using the quotient rule for derivatives, we get: \[\frac{\partial h}{\partial w} = \frac{x y (\frac{\partial}{\partial w} (wz)) - wz (\frac{\partial}{\partial w} (xy))}{(xy)^2}\] Now, since x, y, and z are constants with respect to w, their derivatives with respect to w are all zero. The derivative of wz with respect to w is z, so we have: \[\frac{\partial h}{\partial w} = \frac{x y (z) - 0}{(xy)^2} = \frac{z}{xy}\]
03

Find the partial derivative with respect to x

Now we will find the partial derivative of h with respect to x, treating w, y, and z as constants: $$\frac{\partial h}{\partial x}$$ Using the quotient rule for derivatives, we get: \[\frac{\partial h}{\partial x} = \frac{xy (\frac{\partial}{\partial x} (wz)) - wz (\frac{\partial}{\partial x} (xy))}{(xy)^2}\] Now, since w, y, and z are constants with respect to x, their derivatives with respect to x are all zero. The derivative of xy with respect to x is y, so we have: \[\frac{\partial h}{\partial x} = \frac{0 - wz (y)}{(xy)^2} = -\frac{wz}{x^2 y}\]
04

Find the partial derivative with respect to y

Next, we will find the partial derivative of h with respect to y, treating w, x, and z as constants: $$\frac{\partial h}{\partial y}$$ Using the quotient rule for derivatives, we get: \[\frac{\partial h}{\partial y} = \frac{xy (\frac{\partial}{\partial y} (wz)) - wz (\frac{\partial}{\partial y} (xy))}{(xy)^2}\] Now, since w, x, and z are constants with respect to y, their derivatives with respect to y are all zero. The derivative of xy with respect to y is x, so we have: \[\frac{\partial h}{\partial y} = \frac{0 - wz (x)}{(xy)^2} = -\frac{wxz}{y^2 x}\]
05

Find the partial derivative with respect to z

Finally, we will find the partial derivative of h with respect to z, treating w, x, and y as constants: $$\frac{\partial h}{\partial z}$$ Using the quotient rule for derivatives, we get: \[\frac{\partial h}{\partial z} = \frac{xy (\frac{\partial}{\partial z} (wz)) - wz (\frac{\partial}{\partial z} (xy))}{(xy)^2}\] Now, since w, x, and y are constants with respect to z, their derivatives with respect to z are all zero. The derivative of wz with respect to z is w, so we have: \[\frac{\partial h}{\partial z} = \frac{xy (w) - 0}{(xy)^2} = \frac{w}{xy}\]
06

Write the final answer

The first partial derivatives of the given function are: \[\frac{\partial h}{\partial w} = \frac{z}{xy}, \quad \frac{\partial h}{\partial x} = -\frac{wz}{x^2 y}, \quad \frac{\partial h}{\partial y} = -\frac{wxz}{y^2 x}, \quad \frac{\partial h}{\partial z} = \frac{w}{xy}\]

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Key Concepts

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

Multivariable Calculus
Multivariable calculus is a fascinating extension of single-variable calculus. While single-variable calculus deals with functions that have a single input and output, multivariable calculus allows us to work with functions that have multiple inputs. These inputs are often represented as variables, such as \(w\), \(x\), \(y\), and \(z\) in our example.

The core idea is that each variable could change, and each has an impact on the outcome. Just imagine adjusting each knob on a soundboard, where each knob represents a different variable. Each change alters the sound just like changing each variable alters the function's result in multivariable calculus.
  • Partial derivatives are used to understand the impact of one specific variable on the function while keeping all other variables constant.
  • This involves treating other variables as constants and differentiating with respect to the interested variable.
  • Multivariable calculus allows us to analyze surfaces and curves in higher-dimensional spaces.
By understanding how each variable influences the function individually, we can gain deeper insights into complex systems that rely on multiple factors.
Quotient Rule
The quotient rule is a crucial tool in calculus, especially when dealing with derivatives of functions that involve division. If you remember how to differentiate simple functions, the quotient rule helps you tackle more complicated ones where one function is divided by another.

In our example, we have a function \( h(w, x, y, z) = \frac{wz}{xy} \). The numerator \( wz \) and the denominator \( xy \) are both functions of several variables.
  • The quotient rule states: if you have a function \( \frac{f(x)}{g(x)} \), the derivative is \( \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \).
  • Each of the terms in the rule leverages the power of differentiation. The rule requires you to differentiate both the numerator and the denominator separately.
  • It's essential because it helps in analyzing functions that are not easily simplifiable into a basic derivative form.
Understandably, this can feel complex, but remember it as a step-by-step pattern: differentiate, multiply, and subtract, then divide by the square of the denominator.
Differentiation
Differentiation is a fundamental concept in calculus focused on finding the derivative of a function. The derivative essentially measures how a function changes as its input changes. Think of it as finding the slope of a tangent line to the curve described by the function.

When we talk about concepts like partial derivatives in our original exercise, we're using differentiation to explore how changes in one specific variable affect the overall function.
  • Differentiating with respect to each variable individually helps us understand high-dimensional changes in multivariable functions.
  • Partial derivatives are a type of differentiation where you hold some variables constant while varying others.
  • These derivatives ultimately allow us to build models of change for complex systems or to optimize them.
Differentiation opens the door to predicting how a system behaves by providing insights into rates of change and how different components interact within mathematical models.

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Most popular questions from this chapter

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Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The planes tangent to the cylinder \(x^{2}+y^{2}=1\) in \(R^{3}\) all have the form \(a x+b z+c=0\) b. Suppose \(w=x y / z,\) for \(x>0, y>0,\) and \(z>0 .\) A decrease in \(z\) with \(x\) and \(y\) fixed results in an increase in \(w\) c. The gradient \(\nabla F(a, b, c)\) lies in the plane tangent to the surface \(F(x, y, z)=0\) at \((a, b, c)\)

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