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Chapter 3: Problem 45

Find \(y^{\prime \prime}\) for the following functions. $$y=\cot x$$

Short Answer

Expert verified
Answer: The second derivative of the function \(y=\cot x\) is \(y^{\prime \prime} = 2(\csc^2 x)^2\sin 2x\).

Step by step solution

01

Find the first derivative \(y^{\prime}\)

Recall that \(y = \cot x = \frac{\cos x}{\sin x}\). To find the first derivative, we can use the quotient rule for differentiating two functions: $$y^{\prime} = \frac{(\frac{d(\cos x)}{dx})\sin x - \cos x \frac{d(\sin x)}{dx}}{(\sin x)^2}.$$ Now, we find the derivatives of \(\sin x\) and \(\cos x\): $$\frac{d(\cos x)}{dx} = -\sin x$$ $$\frac{d(\sin x)}{dx} = \cos x$$ Plug these values back into the quotient rule equation.
02

Simplify the first derivative \(y^{\prime}\)

Substitute the computed derivative values into the derivative formula and simplify: $$y^{\prime} = \frac{(-\sin x)\sin x - \cos x (\cos x)}{(\sin x)^2}$$ $$y^{\prime} = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x}$$ Since \(\sin^2 x + \cos^2 x = 1\), we can simplify the first derivative: $$y^{\prime} = \frac{-1}{\sin^2 x}$$ $$y^{\prime} = -\csc^2 x$$
03

Find the second derivative \(y^{\prime \prime}\)

Now, we will find the second derivative \(y^{\prime \prime}\) by differentiating the first derivative with respect to \(x\): $$y^{\prime \prime} = \frac{d(-\csc^2 x)}{dx}$$
04

Use trigonometric identities and differentiation rules

Recall that \(\csc x = \frac{1}{\sin x}\). Then, \(y^{\prime} = -\frac{1}{(\sin^2 x)}\). Using the chain rule of differentiation: $$y^{\prime \prime} = \frac{d(-\frac{1}{\sin^2 x})}{dx} = \frac{d(\frac{-1}{u})}{du} \frac{du}{dx}$$ with \(u = \sin^2 x\). So, we have: $$y^{\prime \prime} = \frac{2}{u^2} \frac{d(\sin^2 x)}{dx}$$ Now, find \(\frac{d(\sin^2 x)}{dx}\) using the chain rule: $$\frac{d(\sin^2 x)}{dx}= 2\sin x \frac{d(\sin x)}{dx} = 2\sin x \cos x = \sin 2x$$ Now plug the value back into equation to find \(y^{\prime \prime}\):
05

Simplify the second derivative \(y^{\prime \prime}\)

Replace the value of the derivative back into the formula for \(y^{\prime \prime}\) and simplify: $$y^{\prime \prime} = \frac{2}{u^2}\sin 2x = \frac{2}{(\sin^2 x)^2}\sin 2x = 2(\csc^2 x)^2\sin 2x$$ Therefore the second derivative \(y^{\prime \prime}\) of the function \(y=\cot x\) is: $$y^{\prime \prime} = 2(\csc^2 x)^2\sin 2x$$

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