91Ó°ÊÓ

We can rewrite the function \(y = \sec x \csc x\) using the reciprocal trigonometric functions as follows: $$y = \frac{1}{\cos x}\cdot\frac{1}{\sin x}$$ We will use this form in the differentiation process.

To find the first derivative \(y^{\prime}\), we will apply the product rule on two functions \(f(x) = \frac{1}{\cos x}\) and \(g(x) = \frac{1}{\sin x}\). The product rule states that: $$(fg)^{\prime} = f^{\prime}g + fg^{\prime}$$ Let's find the derivatives of \(f(x)\) and \(g(x)\) individually. Derivative of \(f(x) = \frac{1}{\cos x}\): $$f^{\prime}(x) = -\frac{\sin x}{\cos^2 x}$$ Derivative of \(g(x) = \frac{1}{\sin x}\): $$g^{\prime}(x) = -\frac{\cos x}{\sin^2 x}$$ Now, let's apply the product rule: $$y^{\prime} = f^{\prime}g + fg^{\prime} = -\frac{\sin x}{\cos^2 x}\cdot\frac{1}{\sin x} + \frac{1}{\cos x}\cdot-\frac{\cos x}{\sin^2 x}$$ Simplify the expression to get: $$y^{\prime} = -\frac{1}{\cos^2 x} - \frac{\cos x}{\sin^2 x}$$

Now that we have \(y^{\prime}\), let's find the second derivative \(y^{\prime\prime}\). As we have a sum of two fractions, let's differentiate each part individually and then sum up the results. First, let's find the derivative of \(-\frac{1}{\cos^2 x}\): $$\frac{d}{dx}\left( -\frac{1}{\cos^2 x} \right) = \frac{d}{dx}\left( -\sec^2 x \right) = 2 \sec x\tan x\sec x = 2 \sec^2 x \tan x$$ Now, let's find the derivative of \(-\frac{\cos x}{\sin^2 x}\): $$\frac{d}{dx}\left( -\frac{\cos x}{\sin^2 x} \right) = -\frac{\sin^2 x \cdot (-\sin x) - \cos x \cdot 2\sin x \cdot \cos x}{(\sin^2 x)^2}$$ Simplify this expression: $$\frac{d}{dx}\left( -\frac{\cos x}{\sin^2 x} \right) = \frac{\sin^3 x + 2\sin x\cos^2 x}{\sin^4 x}$$ Finally, sum up the two expressions to get the second derivative \(y^{\prime\prime}\): $$y^{\prime\prime} = 2 \sec^2 x \tan x + \frac{\sin^3 x + 2\sin x\cos^2 x}{\sin^4 x}$$