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Chapter 3: Problem 64

Use the Chain Rule combined with other differentiation rules to find the derivative of the following functions. $$y=\frac{t e^{t}}{t+1}$$

Short Answer

Expert verified
Based on the given step-by-step solution, a short answer question can be as follows: Question: Given the function $$y=\frac{te^t}{t+1}$$. Find the derivative $$\frac{dy}{dt}$$. Answer: $$\frac{dy}{dt}=\frac{e^t(t^2 + t)}{(t+ 1)^2}$$

Step by step solution

01

Identify the main components of the function

The given function can be written as: $$y = \frac{u(t)}{v(t)}$$ where $$u(t) = t e^t$$ and $$v(t) = t + 1$$ Now, we will find the derivatives of these individual functions.
02

Differentiate u(t)

To find $$\frac{du}{dt}$$, we'll use the product rule since it is a product of two functions, t and $$e^t$$: Product rule: $$(uv)' = u'v + uv'$$ Here, $$u = t$$ and $$v = e^t$$ Now, differentiate u and v: $$\frac{du}{dt} = 1$$ $$\frac{dv}{dt} = e^t$$ Apply the product rule: $$\frac{d}{dt}(t e^t) = 1 \cdot e^t + t \cdot e^t = e^t(1 + t)$$ So, $$\frac{du}{dt} = e^t(1 + t)$$
03

Differentiate v(t)

To find $$\frac{dv}{dt}$$, we'll take the derivative of v(t): $$v(t) = t + 1$$ The derivative of t is 1 and the derivative of a constant is 0, so: $$\frac{dv}{dt} = 1 + 0 = 1$$
04

Apply the Quotient Rule

Now we have all the required components to apply the Quotient Rule. The Quotient Rule states that if we have a function y = u/v, then the derivative of y with respect to t is given by: $$(y)' = \frac{u'v - uv'}{v^2}$$ Applying the Quotient Rule using $$\frac{du}{dt}$$(step2) and $$\frac{dv}{dt}$$(step 3): $$\frac{dy}{dt}=\frac{(e^t(1 + t))(t + 1)- (t e^t)(1)}{(t + 1)^2}$$ Now, simplify the expression: $$\frac{dy}{dt}=\frac{e^t(t^2+ 2t + 1)-t e^t}{(t+ 1)^2}$$ $$\frac{dy}{dt}=\frac{e^t(t^2 + t)}{(t+ 1)^2}$$ So, the derivative of the given function is: $$\frac{dy}{dt}=\frac{e^t(t^2 + t)}{(t+ 1)^2}$$

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Most popular questions from this chapter

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

A lighthouse stands 500 m off of a straight shore, the focused beam of its light revolving four times each minute. As shown in the figure, \(P\) is the point on shore closest to the lighthouse and \(Q\) is a point on the shore 200 m from \(P\). What is the speed of the beam along the shore when it strikes the point \(Q ?\) Describe how the speed of the beam along the shore varies with the distance between \(P\) and \(Q\). Neglect the height of the lighthouse.

An observer stands \(20 \mathrm{m}\) from the bottom of a 10 -m-tall Ferris wheel on a line that is perpendicular to the face of the Ferris wheel. The wheel revolves at a rate of \(\pi\) rad/min and the observer's line of sight with a specific seat on the wheel makes an angle \(\theta\) with the ground (see figure). Forty seconds after that seat leaves the lowest point on the wheel, what is the rate of change of \(\theta ?\) Assume the observer's eyes are level with the bottom of the wheel.

One of the Leibniz Rules One of several Leibniz Rules in calculus deals with higher-order derivatives of products. Let \((f g)^{(n)}\) denote the \(n\) th derivative of the product \(f g,\) for \(n \geq 1\) a. Prove that \((f g)^{(2)}=f^{\prime \prime} g+2 f^{\prime} g^{\prime}+f g^{\prime \prime}\) b. Prove that, in general,$$(f g)^{(n)}=\sum_{k=0}^{n}\left(\begin{array}{l} n \\\k\end{array}\right) f^{(k)} g^{(n-k)}$$ where \(\left(\begin{array}{l}n \\\ k\end{array}\right)=\frac{n !}{k !(n-k) !}\) are the binomial coefficients. c. Compare the result of (b) to the expansion of \((a+b)^{n}\).

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