Find $y^{\prime \prime}$ if $y=4 \sec x$

Step 1 ：We are given the function \(y = 4 \sec x\). We need to find the second derivative of this function.

Step 2 ：First, we rewrite the secant function in terms of cosine: \(y = \frac{4}{\cos x}\).

Step 3 ：Next, we find the first derivative of \(y\) using the quotient rule. The quotient rule states that the derivative of \(\frac{f(x)}{g(x)}\) is \(\frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\). Here, \(f(x) = 4\) and \(g(x) = \cos x\). The derivative of \(f(x)\) is 0 and the derivative of \(g(x)\) is \(-\sin x\).

Step 4 ：Applying the quotient rule, we find that the first derivative of \(y\) is \(y' = \frac{4 \sin x}{\cos^2 x}\).

Step 5 ：We then find the second derivative by differentiating the first derivative. Again using the quotient rule, we find that the second derivative of \(y\) is \(y'' = \frac{8 \sin^2 x}{\cos^3 x} + \frac{4}{\cos x}\).

Step 6 ：\(\boxed{y'' = \frac{8 \sin^2 x}{\cos^3 x} + \frac{4}{\cos x}}\) is the final answer.