Question

In Exercises \(75-80\), evaluate the derivative of the function at the indicated point. Use a graphing utility to verify your result. \(\frac{\text { Function }}{y=37-\sec ^{3}(2 x)} \quad \frac{\text { Point }}{(0,36)}\)

Short answer

The derivative of the function \(y = 37 - \sec^{3}(2x)\) at the point (0,36) is -6. This result is verified using a graphing utility.

Step-by-step solution

Find the Derivative of the Function

Start by finding the derivative of \(y = 37 - \sec^{3}(2x)\). The derivative of a constant is zero, and the derivative of \(-\sec^{3}(2x)\) can be found using the chain rule. If \(f(x) = h(g(x))\), where h(x) = \(-\sec^{3}(x)\) and g(x) = 2x, we have \(f'(x) = h'(g(x)) \cdot g'(x)\). Thus, the derivative of \(-\sec^{3}(2x)\) is \(-3\sec^{2}(2x) \cdot \tan(2x) \cdot 2\). The overall derivative is then \(y' = 0 - 3\sec^{2}(2x) \cdot \tan(2x) \cdot 2 or -6\sec^{2}(2x) \cdot \tan(2x)\).

Evaluate the Derivative at the Given Point

Next, evaluate the derivative from Step 1, -6\sec^{2}(2x) \cdot \tan(2x), at the given point (0,36). The x-coordinate at this point is zero, so we substitute 0 into the equation: -6\sec^{2}(2*0) \cdot \tan(2*0) = -6.

Verify the Result Using a Graphing Utility

The result can be verified using a graphing utility. Simply, plot the initial function \(y = 37 - \sec^{3}(2x)\) and its derivative found in Step 1, then find the slope of the tangent line at the point (0,36). It should match the computed derivative at this point, which is -6.