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Chapter 12: Problem 23

Find the first partial derivatives of the following functions. $$s(y, z)=z^{2} \tan y z$$

Short Answer

Expert verified
Answer: The first partial derivatives of the function $$s(y, z)$$ are: $$\frac{\partial s}{\partial y} = z^3\sec^2(yz)$$ $$\frac{\partial s}{\partial z} = 2z\tan(yz) + yz^2\sec^2(yz)$$

Step by step solution

01

Find Partial Derivative with Respect to y #

To find the partial derivative of $$s(y,z)$$ with respect to y, we will differentiate the function treating z as constant. We can use the product rule: $$(u \cdot v)'=u'v + uv'$$ Let u = $$z^2$$ and v = $$\tan (yz)$$. The derivative of u with respect to y is: $$\frac{\partial u}{\partial y} = \frac{\partial (z^2)}{\partial y} = 0 $$ The derivative of v with respect to y is: $$\frac{\partial v}{\partial y} = \frac{\partial (\tan(yz))}{\partial y}$$ Using the chain rule and noting that the derivative of the tangent function is $$\sec^2(x)$$, we get: $$\frac{\partial v}{\partial y} = \sec^2(yz)\frac{\partial (yz)}{\partial y} = \sec^2(yz)(z)$$ Now we can apply the product rule to get the partial derivative of s with respect to y: $$\frac{\partial s}{\partial y} = 0 \cdot \tan(yz) + z^2\sec^2(yz)(z) = z^3\sec^2(yz)$$
02

Find Partial Derivative with Respect to z #

To find the partial derivative of $$s(y,z)$$ with respect to z, we will differentiate the function, treating y as constant. Using product rule with u = $$z^2$$ and v = $$\tan (yz)$$ again, the derivative of u with respect to z is: $$\frac{\partial u}{\partial z} = \frac{\partial (z^2)}{\partial z} = 2z$$ The derivative of v with respect to z is: $$\frac{\partial v}{\partial z} = \frac{\partial (\tan(yz))}{\partial z}$$ Using the chain rule, we get: $$\frac{\partial v}{\partial z} = \sec^2(yz)\frac{\partial (yz)}{\partial z} = \sec^2(yz)(y)$$ Now we can apply the product rule to get the partial derivative of s with respect to z: $$\frac{\partial s}{\partial z} = (2z) \cdot \tan(yz) + z^2(\sec^2(yz)(y)) = 2z\tan(yz) + yz^2\sec^2(yz)$$ Finally, the first partial derivatives of the function $$s(y,z)$$ are: $$\frac{\partial s}{\partial y} = z^3\sec^2(yz)$$ $$\frac{\partial s}{\partial z} = 2z\tan(yz) + yz^2\sec^2(yz)$$

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Key Concepts

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

Product Rule
The product rule is a fundamental tool in calculus, especially when dealing with functions that are the product of two or more separate functions.
When you have a function that is expressed as the product of two others, such as \( u(x) \ imes v(x) \), the product rule helps you differentiate it.
It can be formulated as: \((u \ imes v)' = u'v + uv'\). This rule basically states that you can find the derivative of a product by differentiating each part separately and then combining them using the sum of their products in a specific way.
  • \( u' \) represents the derivative of the first function \( u \).
  • \( v' \) is the derivative of the second function \( v \).
  • You multiply the derivative of the first function by the second function, and the first function by the derivative of the second, then add these two results together.
In the given problem, the partial derivative of \( s(y,z) = z^2 \tan(yz) \) uses the product rule to differentiate \( u = z^2 \) and \( v = \tan(yz) \) separately.
Chain Rule
The chain rule is another vital tool in calculus, especially when your function involves a composition of two or more functions.
It essentially allows you to differentiate functions of the form \( f(g(x)) \) by finding the derivative of the outer function evaluated at the inner function and multiplying by the derivative of the inner function:
\( (f(g(x)))' = f'(g(x)) \times g'(x) \).In multivariable calculus, where functions are more complex, the chain rule is similarly used to manage the derivative involving nested functions or variables interacting.
In the solution, the chain rule is applied in finding the derivative of \( \tan(yz) \), where \( yz \) is seen as the inner function.
  • The derivative of \( \tan(x) \) is \( \sec^2(x) \).
  • Thus, for \( \tan(yz) \), we multiply \( \sec^2(yz) \) by the derivative of \( yz \) with respect to the variable being differentiated (either \( y \) or \( z \)).
Multivariable Calculus
Multivariable calculus extends the principles of calculus to functions of multiple variables.
While single-variable calculus deals with functions involving one independent variable, multivariable calculus allows for several independent variables and studies how they interact.This field introduces several new concepts such as partial derivatives, which measure how a function changes as one of the variables changes, while keeping the others constant.
  • For the function \( s(y,z) = z^2 \tan(yz) \), finding the partial derivative with respect to \( y \) involves treating \( z \) as a constant.
  • Conversely, finding the partial derivative with respect to \( z \) treats \( y \) as a constant.
Using partial derivatives, one can understand how each variable influences the function's behavior, an essential tool in fields where multiple factors are at play, such as physics, engineering, and economics.

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Most popular questions from this chapter

A clothing company makes a profit of \(\$ 10\) on its long-sleeved T-shirts and \(\$ 5\) on its short-sleeved T-shirts. Assuming there is a \(\$ 200\) setup cost, the profit on \(\mathrm{T}\) -shirt sales is \(z=10 x+5 y-200,\) where \(x\) is the number of long-sleeved T-shirts sold and \(y\) is the number of short-sleeved T-shirts sold. Assume \(x\) and \(y\) are nonnegative. a. Graph the plane that gives the profit using the window $$ [0,40] \times[0,40] \times[-400,400] $$ b. If \(x=20\) and \(y=10,\) is the profit positive or negative? c. Describe the values of \(x\) and \(y\) for which the company breaks even (for which the profit is zero). Mark this set on your graph.

Suppose that in a large group of people, a fraction \(0 \leq r \leq 1\) of the people have flu. The probability that in \(n\) random encounters you will meet at least one person with flu is \(P=f(n, r)=1-(1-r)^{n} .\) Although \(n\) is a positive integer, regard it as a positive real number. a. Compute \(f_{r}\) and \(f_{n}.\) b. How sensitive is the probability \(P\) to the flu rate \(r ?\) Suppose you meet \(n=20\) people. Approximately how much does the probability \(P\) increase if the flu rate increases from \(r=0.1\) to \(r=0.11(\text { with } n \text { fixed }) ?\) c. Approximately how much does the probability \(P\) increase if the flu rate increases from \(r=0.9\) to \(r=0.91\) with \(n=20 ?\) d. Interpret the results of parts (b) and (c).

The flow of heat along a thin conducting bar is governed by the one- dimensional heat equation (with analogs for thin plates in two dimensions and for solids in three dimensions) $$\frac{\partial u}{\partial t}=k \frac{\partial^{2} u}{\partial x^{2}},$$ where \(u\) is a measure of the temperature at a location \(x\) on the bar at time t and the positive constant \(k\) is related to the conductivity of the material. Show that the following functions satisfy the heat equation with \(k=1\). $$u(x, t)=e^{-t}(2 \sin x+3 \cos x)$$

Consider the following equations of quadric surfaces. a. Find the intercepts with the three coordinate axes, when they exist. b. Find the equations of the x y-, x z^{-}, \text {and } y z-\text {traces, when they exist. c. Sketch a graph of the surface. $$-\frac{x^{2}}{6}-24 y^{2}+\frac{z^{2}}{24}-6=0$$

Given the production function \(P=f(K, L)=K^{a} L^{1-a}\) and the budget constraint \(p K+q L=B,\) where \(a, p, q,\) and \(B\) are given, show that \(P\) is maximized when \(K=a B / p\) and \(L=(1-a) B / q\).
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