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Chapter 3: Problem 67

Use the Chain Rule combined with other differentiation rules to find the derivative of the following functions. $$y=(p+\pi)^{2} \sin p^{2}$$

Short Answer

Expert verified
A) Chain Rule B) Product Rule C) Quotient Rule D) Chain Rule and Product Rule E) Chain Rule, Quotient Rule, and Product Rule

Step by step solution

01

Differentiate \((p+\pi)^{2}\)

To differentiate \((p+\pi)^{2}\), we will use the Chain Rule, as follows: $$\frac{d}{dp}(p+\pi)^{2} = 2(p+\pi)\cdot \frac{d}{dp}(p+\pi)$$ Since \(\pi\) is a constant, its derivative is zero. So, we have: $$\frac{d}{dp}(p+\pi)^{2} = 2(p+\pi)\cdot 1 = 2(p+\pi)$$
02

Differentiate \(\sin p^{2}\)

To differentiate \(\sin p^{2}\), we will also use the Chain Rule, as follows: $$\frac{d}{dp}\sin p^{2} = (\cos p^{2})\cdot \frac{d}{dp}p^{2}$$ To find the derivative of \(p^{2}\), we have: $$\frac{d}{dp}p^{2} = 2p$$ So, the derivative of \(\sin p^{2}\) is: $$\frac{d}{dp}\sin p^{2} = \cos p^{2}\cdot 2p = 2p\cos p^{2}$$
03

Apply the Product Rule

Now we can apply the Product Rule using the derivatives we found in Steps 1 and 2: $$\frac{dy}{dp} = (2(p+\pi))\cdot (\sin p^{2}) + ((p+\pi)^{2})\cdot (2p\cos p^{2})$$ So, the derivative of the given function is: $$\frac{dy}{dp} = 2(p+\pi)\sin p^{2} + 2p(p+\pi)^{2}\cos p^{2}$$

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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 essential in calculus when dealing with the differentiation of functions that are multiplied together. If you have two functions, say \( u(x) \) and \( v(x) \), their product \( u(x) \cdot v(x) \) can be differentiated as follows:
  • Let \( u(x) = (p+\pi)^2 \) and \( v(x) = \sin p^2 \) for our exercise.
  • According to the Product Rule: \( \frac{d}{dp}(u \cdot v) = u'v + uv' \).
This means we need to find the derivative of both \( u \) and \( v \) separately and then apply the rule:
  • The derivative of \( u = (p+\pi)^2 \) is \( 2(p+\pi) \), as found using the Chain Rule.
  • The derivative of \( v = \sin p^2 \) is \( 2p \cos p^2 \), also using the Chain Rule.
By applying these derivatives, we combine them as \( (2(p+\pi)\sin p^2) + ((p+\pi)^2 \cdot 2p \cos p^2) \), giving us the final differentiated expression.
Differentiation
Differentiation is a fundamental concept in calculus used to find the rate at which a function is changing at any given point. It involves taking the derivative of a function, which provides the slope of the tangent line at a particular point. In our exercise, differentiation helps us find how the function \( y=(p+\pi)^{2} \sin p^{2} \) changes concerning \( p \).
  • The Chain Rule and Product Rule are often used in combination for such tasks.
  • Each term must be differentiated separately, respecting constants like \( \pi \), which have a derivative of zero.
Here, we use differentiation to manage both parts of the function:
  • When differentiating \( (p+\pi)^2 \), the Chain Rule simplifies the process.
  • \( \sin p^2 \) shows how chain differentiation works with trigonometric functions.
This combination ensures we accurately find how each part of the function changes.
Trigonometric Functions
Trigonometric functions, like \( \sin \) and \( \cos \), are important in calculus for modeling periodic phenomena. In differentiation, they require special attention, especially when they are involved in composite functions, as they are sensitive to changes in their input.
  • Consider \( \sin p^2 \) from our exercise. To differentiate this, the Chain Rule is essential.
  • We first take the derivative of the outer function \( \sin \, x \). For \( \sin \, x \), the derivative is \( \cos \, x \).
  • Next, we differentiate the inner function \( p^2 \), yielding \( 2p \).
Thus, combining these steps gives us \( 2p \cos p^2 \). Trigonometric functions can change rapidly, and their derivatives provide insights into these changes, depicting how their wave-like patterns evolve over different inputs.

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Most popular questions from this chapter

Suppose \(f\) is differentiable on an interval containing \(a\) and \(b\), and let \(P(a, f(a))\) and \(Q(b, f(b))\) be distinct points on the graph of \(f\). Let \(c\) be the \(x\) -coordinate of the point at which the lines tangent to the curve at \(P\) and \(Q\) intersect, assuming that the tangent lines are not parallel (see figure). a. If \(f(x)=x^{2},\) show that \(c=(a+b) / 2,\) the arithmetic mean of \(a\) and \(b\), for real numbers \(a\) and \(b\) b. If \(f(x)=\sqrt{x}\), show that \(c=\sqrt{a b}\), the geometric mean of \(a\) and \(b\), for \(a > 0\) and \(b > 0\) c. If \(f(x)=1 / x,\) show that \(c=2 a b /(a+b),\) the harmonic mean of \(a\) and \(b,\) for \(a > 0\) and \(b > 0\) d. Find an expression for \(c\) in terms of \(a\) and \(b\) for any (differentiable) function \(f\) whenever \(c\) exists.

Let \(C(x)\) represent the cost of producing \(x\) items and \(p(x)\) be the sale price per item if \(x\) items are sold. The profit \(P(x)\) of selling x items is \(P(x)=x p(x)-C(x)\) (revenue minus costs). The average profit per item when \(x\) items are sold is \(P(x) / x\) and the marginal profit is dP/dx. The marginal profit approximates the profit obtained by selling one more item given that \(x\) items have already been sold. Consider the following cost functions \(C\) and price functions \(p\). a. Find the profit function \(P\). b. Find the average profit function and marginal profit function. c. Find the average profit and marginal profit if \(x=a\) units are sold. d. Interpret the meaning of the values obtained in part \((c)\). $$\begin{aligned} &C(x)=-0.04 x^{2}+100 x+800, p(x)=200-0.1 x,\\\ &\bar{a}=1000 \end{aligned}$$

The lateral surface area of a cone of radius \(r\) and height \(h\) (the surface area excluding the base) is \(A=\pi r \sqrt{r^{2}+h^{2}}\) a. Find \(d r / d h\) for a cone with a lateral surface area of $$ A=1500 \pi $$ b. Evaluate this derivative when \(r=30\) and \(h=40\)

Logistic growth Scientists often use the logistic growth function \(P(t)=\frac{P_{0} K}{P_{0}+\left(K-P_{0}\right) e^{-r_{d}}}\) to model population growth, where \(P_{0}\) is the initial population at time \(t=0, K\) is the carrying capacity, and \(r_{0}\) is the base growth rate. The carrying capacity is a theoretical upper bound on the total population that the surrounding environment can support. The figure shows the sigmoid (S-shaped) curve associated with a typical logistic model. World population (part 1 ) The population of the world reached 6 billion in \(1999(t=0)\). Assume Earth's carrying capacity is 15 billion and the base growth rate is \(r_{0}=0.025\) per year. a. Write a logistic growth function for the world's population (in billions) and graph your equation on the interval \(0 \leq t \leq 200\) using a graphing utility. b. What will the population be in the year 2020? When will it reach 12 billion?

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{x \rightarrow 2} \frac{5^{x}-25}{x-2}$$
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