Question

Evaluating a Derivative In Exercises \(65-72,\) find and evaluate the derivative of the function at the given point. Use a graphing utility to verify your result. $$ y=26-\sec ^{3} 4 x, \quad(0,25) $$

Short answer

The derivative of the function \(y=26-\sec^3 4x\) at the point (0,25) is 0.

Step-by-step solution

Find the derivative

Firstly, find the derivative of the given function. The function is \(y = 26 - \sec^3 4x \). The derivative of a constant is zero and the derivative of \( \sec x \) is \( \sec x \tan x \), but we also need to consider the chain rule because of the \(4x\) inside the secant function, which will multiply the derivative by 4. With this in consideration, we differentiate to get \( y' = 0 - 3 \sec^2 4x * \sec 4x \tan 4x * 4 \), which simplifies to \( y' = -12 \sec^2 4x * \sec 4x \tan 4x \).

Evaluate the derivative at the given point

Next, plug the x value from the given point (0,25) into the derivative equation. This gives us \( y'(0) = -12 \sec^2 (4*0) * \sec (4*0) \tan (4*0) = -12 \sec^2 (0) * \sec (0) \tan (0) = -12(1)^2*(1)^2*(0) = 0 \).

Verify the result using a graphing utility

To confirm this result, use a graphing utility to graph the function and its derivative. The slope of the tangent line to the function at the point (0,25) should be the same as the derivative at that point, which is 0.

Key concepts

Derivative of Secant Function

Understanding the derivative of the secant function is essential for tackling a variety of calculus problems. The secant function, denoted as \( \sec(x) \) , is one of the trigonometric functions and is defined as the reciprocal of the cosine function; that is, \( \sec(x) = \frac{1}{\cos(x)} \). When taking the derivative of \( \sec(x) \) with respect to \( x \), it is important to remember the result is \( \sec(x)\tan(x) \).

However, when the secant function is raised to a power, such as \( \sec^3(x) \), or when it is multiplied by another function of \( x \), as in \( \sec^3(4x) \), we must apply the power rule in conjunction with the chain rule to correctly differentiate the function. For instance, the derivative of \( \sec^3(x) \) is \( 3\sec^2(x)\sec(x)\tan(x) \), which incorporates the power rule. But for \( \sec^3(4x) \) we multiply by the derivative of \( 4x \) due to the chain rule, resulting in the derivative \( 3\sec^2(4x)\sec(4x)\tan(4x)\times 4 \).

It is pivotal to approach these derivatives systematically, ensuring each step follows from the application of the correct rules of differentiation.

Chain Rule in Calculus

The chain rule is a fundamental tool in calculus that we use when differentiating compositions of functions. It states that to find the derivative of a composite function, you multiply the derivative of the outer function by the derivative of the inner function. This can seem confusing, but a step-by-step approach makes it manageable.

Let's consider a function in the form of \( f(g(x)) \), where \( f \) is the outer function and \( g \) is the inner function. The chain rule tells us that the derivative, \( f'(g(x)) \), is, in fact, \( f'(g(x))\times g'(x) \). In the context of our exercise, where we had to differentiate \( \sec^3(4x) \), the outer function is \( \sec^3(u) \) where \( u = 4x \) is the inner function. By applying the chain rule, we first differentiate \( \sec^3(u) \) with respect to \( u \) and then multiply by the derivative of \( 4x \) with respect to \( x \), which is \( 4 \). The chain rule ensures that we correctly account for every layer of composition in our function when finding its derivative.

Graphing Utility Verification

After performing complex calculations, it is crucial to validate our results. One effective method of verification is using a graphing utility. Graphing utilities are powerful tools that provide visual representation of functions and their derivatives, which are particularly handy to confirm the accuracy of our calculated derivatives.

Using a graphing utility, we can plot \( y = 26 - \sec^3(4x) \) and its derivative. The slope of the tangent line at any given point on our function's graph should correspond to the derivative at that same point. In our exercise, we evaluated the derivative at point \( (0, 25) \) and found it to be \( 0 \) - this indicates that the tangent line at this point is horizontal.

When you input the function into the graphing utility and examine the graph at \( x = 0 \), you expect to see a horizontal line touching the curve at \( y = 25 \). If the visual matches our expectation, then the graphing utility has verified the result of our derivative calculation. It is an intersection of technology with calculus that bolsters our confidence in the results of what can often be intricate mathematical work.