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Chapter 3: Problem 65

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
Question: Find the derivative of the function \(y=(p+\pi)^{2} \sin p^{2}\) using the Chain Rule and other differentiation rules. Answer: The derivative of the function is \(\frac{dy}{dp} = 2(p+\pi) \sin p^{2} + 2p(p+\pi)^{2} \cos p^{2}\).

Step by step solution

01

Identify the Product Rule

In this scenario, we have to differentiate the function \(y=(p+\pi)^{2} \sin p^{2}\). Notice that the function can be written as a product of two functions: \(f(p) = (p+\pi)^{2}\) and \(g(p) = \sin p^{2}\). The Product Rule states that: $$\frac{dy}{dp} = \frac{df(p)}{dp} \cdot g(p) + f(p) \cdot \frac{dg(p)}{dp}$$ We will now differentiate \(f(p)\) and \(g(p)\) before substituting their derivatives into the Product Rule formula.
02

Differentiate \((p+\pi)^{2}\) using Chain Rule

The Chain Rule states that: $$\frac{d}{dp}(h(g(p))) = h'(g(p)) \cdot g'(p)$$ Let \(h(x) = x^2\) and \(g(p) = p+\pi\). Then, \(h'(x) = 2x\) and \(g'(p) = 1\). Applying the Chain Rule, we get: $$\frac{d}{dp} (p+\pi)^{2} = 2(p+\pi) \cdot 1$$
03

Differentiate \(\sin p^{2}\) using Chain Rule

Let \(h(x) = \sin x\) and \(g(p) = p^2\). Then, \(h'(x) = \cos x\) and \(g'(p) = 2p\). Applying the Chain Rule, we get: $$\frac{d}{dp} \sin p^{2} = \cos p^{2} \cdot 2p$$
04

Apply the Product Rule

Now we have the derivatives of \(f(p)\) and \(g(p)\), so we can apply the Product Rule as follows: \begin{align*} \frac{dy}{dp} &= \frac{d}{dp}(p+\pi)^{2} \cdot \sin p^{2} + (p+\pi)^{2} \cdot \frac{d}{dp} \sin p^{2} \\ &= 2(p+\pi) \cdot \sin p^{2} + (p+\pi)^{2} \cdot (2p \cos p^{2}) \end{align*} After simplification, we obtain the derivative of the given function: $$\frac{dy}{dp} = 2(p+\pi) \sin p^{2} + 2p(p+\pi)^{2} \cos p^{2}$$

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Most popular questions from this chapter

Gravitational force The magnitude of the gravitational force between two objects of mass \(M\) and \(m\) is given by \(F(x)=-\frac{G M m}{x^{2}},\) where \(x\) is the distance between the centers of mass of the objects and \(G=6.7 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\) is the gravitational constant (N stands for newton, the unit of force; the negative sign indicates an attractive force). a. Find the instantaneous rate of change of the force with respect to the distance between the objects. b. For two identical objects of mass \(M=m=0.1 \mathrm{kg},\) what is the instantaneous rate of change of the force at a separation of \(x=0.01 \mathrm{m} ?\) c. Does the instantaneous rate of change of the force increase or decrease with the separation? Explain.

Let $$g(x)=\left\\{\begin{array}{cl} \frac{1-\cos x}{2 x} & \text { if } x \neq 0 \\ a & \text { if } x=0 \end{array}\right.$$ For what values of \(a\) is \(g\) continuous?

Special Quotient Rule In general, the derivative of a quotient is not the quotient of the derivatives. Find nonconstant functions \(f\) and \(g\) such that the derivative of \(f / g\) equals \(f^{\prime} / g^{\prime}\).

Use the following table to find the given derivatives. $$\begin{array}{llllll} x & 1 & 2 & 3 & 4 & 5 \\ \hline f(x) & 5 & 4 & 3 & 2 & 1 \\ f^{\prime}(x) & 3 & 5 & 2 & 1 & 4 \\ g(x) & 4 & 2 & 5 & 3 & 1 \\ g^{\prime}(x) & 2 & 4 & 3 & 1 & 5 \end{array}$$ $$\left.\frac{d}{d x}\left[\frac{x f(x)}{g(x)}\right]\right|_{x=4}$$

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The derivative \(\frac{d}{d x}\left(e^{5}\right)\) equals \(5 \cdot e^{4}\) b. The Quotient Rule must be used to evaluate \(\frac{d}{d x}\left(\frac{x^{2}+3 x+2}{x}\right)\) c. \(\frac{d}{d x}\left(\frac{1}{x^{5}}\right)=\frac{1}{5 x^{4}}\) d. \(\frac{d^{n}}{d x^{n}}\left(e^{3 x}\right)=3^{n} \cdot e^{3 x},\) for any integer \(n \geq 1\)
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