91Ó°ÊÓ

To find the first derivative, let's rewrite the cotangent function as a quotient, \(\cot x = \frac{\cos x}{\sin x}\). Now use the quotient rule, which states \((\frac{u}{v})^{\prime} = \frac{u^{\prime}v - uv^{\prime}}{{v^2}}\), where \(u = \cos x\) and \(v = \sin x\). The derivatives \(u^{\prime}\) and \(v^{\prime}\) are: $$u^{\prime} = -\sin x$$ $$v^{\prime} = \cos x$$ Now, we can apply the quotient rule: $$y^{\prime} = \frac{-\sin x \sin x - \cos x \cos x}{\sin^2 x}$$ Simplify the expression: $$y^{\prime} = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x}$$ As we know, \(\sin^2 x + \cos^2 x = 1\). Therefore, $$y^{\prime} = \frac{-1}{\sin^2 x} = -\csc^2 x$$

Now that we have found the first derivative, \(y^{\prime} = -\csc^2 x\), we can find the second derivative by taking the derivative of \(y^{\prime}\) with respect to x. Recall that \(\csc x = \frac{1}{\sin x}\), therefore \(\csc^2 x = \frac{1}{\sin^2 x}\) and \(y^{\prime} = -\frac{1}{\sin^2 x}\). Using the chain rule, we find the derivative of \(y^{\prime}\) with respect to x: $$y^{\prime \prime} = \frac{d}{dx}\left(-\frac{1}{\sin^2 x}\right)$$ Let \(u = \sin x\), then \(y^{\prime \prime} = \frac{d}{dx}\left(-\frac{1}{u^2}\right)\) and: $$\frac{d}{du} \left(-\frac{1}{u^2}\right) = \frac{2}{u^3}$$ By the chain rule: $$y^{\prime \prime} = \frac{d}{dx} \left(-\frac{1}{u^2}\right) = \frac{2}{u^3} \frac{du}{dx}$$ Now we substitute \(u = \sin x\) and find \(\frac{du}{dx}\): $$\frac{du}{dx} = \frac{d}{dx}(\sin x) = \cos x$$ So, $$y^{\prime \prime} = \frac{2}{\sin^3 x} \cos x = 2\cos x \csc^3 x$$ Thus, the second derivative of the given function \(y = \cot x\) is: $$y^{\prime \prime} = 2\cos x \csc^3 x$$