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Derivatives are a powerful tool in calculus that help us understand how a function changes. For trigonometric functions, specific rules apply. Let's take a closer look at differentiating some common trigonometric functions.- The derivative of the tangent function, \( \tan x \), is \( \sec^2 x \). This derivative arises because the tangent function can be expressed in terms of sine and cosine as \( \frac{\sin x}{\cos x} \), and applying the quotient rule yields \( \sec^2 x \).- The derivative of the cotangent function, \( \cot x \), is \( -\csc^2 x \). The cotangent function is the reciprocal of tangent, which means it can also be expressed in different forms, resulting in this negative derivative.In our problem, we differentiate \( y = \tan x + 3 \cot x \) to find the first derivative, \( y' = \sec^2 x - 3\csc^2 x \). This derivative helps us pinpoint where the function's slope is zero, leading us toward critical points.

Critical points in calculus occur where the derivative of a function is zero or undefined. These points often indicate maximum or minimum values on the graph, making them essential for optimization problems.Here’s how we identify and use critical points:- First, calculate the derivative of the function. In our example, this was \( y' = \sec^2 x - 3\csc^2 x \).- Next, set the derivative equal to zero to find critical points: \( \sec^2 x = 3\csc^2 x \).- Solve the equation to find potential critical values of \( x \). For this problem, solving gives \( \tan^2 x = 3 \), leading to \( \tan x = \sqrt{3} \).- Determine the values of \( x \) within the given interval where these conditions hold. In this case, the critical point is at \( x = \frac{\pi}{3} \).These steps help us to capture where significant changes occur in the function's behavior.

The second derivative test is a method used to determine if a critical point is a minimum, maximum, or a saddle point (neither).Here's how the test works:- Compute the second derivative of the function. For our function, start from \( y' = \sec^2 x - 3\csc^2 x \) and differentiate again to find the second derivative, \( y'' = 2\sec^2 x \tan x + 6\csc^2 x \cot x \).- Plug the critical point into the second derivative. Evaluate it at \( x = \frac{\pi}{3} \).- Check the sign of the second derivative: - If \( y'' > 0 \), the function is concave up and the critical point is a local minimum. - If \( y'' < 0 \), the function is concave down and the critical point is a local maximum.In the current problem, since \( y'' > 0 \) when evaluated, it confirms that the critical point found is indeed a minimum. This method strengthens our confidence in locating the exact nature of significant points on the function's graph.