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Chapter 3: Problem 37

Find the derivative of the following functions. $$y=\frac{\cot x}{1+\csc x}$$

Short Answer

Expert verified
Answer: The derivative of the given function is \(\frac{dy}{dx} = \frac{-\csc^2 x - \csc^3 x + \cot^2 x \csc x}{(1+\csc x)^2}\).

Step by step solution

01

Determine the derivatives of cotangent and cosecant functions

First, we need to express the function \(y\) in terms of sine and cosine. Recall that \(\cot x = \frac{\cos x}{\sin x}\) and \(\csc x = \frac{1}{\sin x}\). So our function becomes: $$y = \frac{\frac{\cos x}{\sin x}}{1+\frac{1}{\sin x}}$$ The derivatives of sine and cosine functions are as follows: $$\frac{d}{dx} \sin x = \cos x \quad \text{and} \quad \frac{d}{dx} \cos x = -\sin x$$ Using these, we can find the derivatives of cotangent and cosecant functions: $$\frac{d}{dx} \cot x = \frac{d}{dx} \left( \frac{\cos x}{\sin x} \right) = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = -\csc^2 x$$ $$\frac{d}{dx} \csc x = \frac{d}{dx} \left( \frac{1}{\sin x} \right) = -\frac{\cos x}{\sin^2 x} = -\cot x \csc x$$
02

Apply the quotient rule to find the derivative of the function

Now we will find the derivative of the function, \(y\), using the quotient rule. The quotient rule states: $$\frac{d}{dx} \frac{u}{v} = \frac{vu' - uv'}{v^2}$$ Let \(u\) be \(\cot x\) and \(v\) be \((1 + \csc x)\). $$\frac{dy}{dx} = \frac{(1+ \csc x)(-\csc^2 x) - \cot x (-\cot x \csc x)}{(1+\csc x)^2}$$ Now simplify the derivative.
03

Simplify the derivative

Simplify the result obtained from the quotient rule: $$\frac{dy}{dx} = \frac{-\csc^2 x - \csc^3 x + \cot^2 x \csc x}{(1+\csc x)^2}$$ Finally, the derivative of the given function is: $$\frac{dy}{dx} = \frac{-\csc^2 x - \csc^3 x + \cot^2 x \csc x}{(1+\csc x)^2}$$

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Most popular questions from this chapter

A woman attached to a bungee cord jumps from a bridge that is \(30 \mathrm{m}\) above a river. Her height in meters above the river \(t\) seconds after the jump is \(y(t)=15\left(1+e^{-t} \cos t\right),\) for \(t \geq 0\). a. Determine her velocity at \(t=1\) and \(t=3\). b. Use a graphing utility to determine when she is moving downward and when she is moving upward during the first 10 s. c. Use a graphing utility to estimate the maximum upward velocity.

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{h \rightarrow 0} \frac{(3+h)^{3+h}-27}{h}$$

Use the properties of logarithms to simplify the following functions before computing \(f^{\prime}(x)\). $$f(x)=\ln \left(\sec ^{4} x \tan ^{2} x\right)$$

A lighthouse stands 500 m off a straight shore and the focused beam of its light revolves 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.

A trough in the shape of a half cylinder has length \(5 \mathrm{m}\) and radius \(1 \mathrm{m}\). The trough is full of water when a valve is opened, and water flows out of the bottom of the trough at a rate of \(1.5 \mathrm{m}^{3} / \mathrm{hr}\) (see figure). (Hint: The area of a sector of a circle of a radius \(r\) subtended by an angle \(\theta\) is \(r^{2} \theta / 2 .\) ) a. How fast is the water level changing when the water level is \(0.5 \mathrm{m}\) from the bottom of the trough? b. What is the rate of change of the surface area of the water when the water is \(0.5 \mathrm{m}\) deep?
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