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. $$ y=\frac{1}{x}+\sqrt{\cos x}, \quad\left(\frac{\pi}{2}, \frac{2}{\pi}\right) $$

Short answer

The derivative of the function at the point \((\frac{\pi}{2}, \frac{2}{\pi})\) is \(-\frac{4}{\pi^2}\) .

Step-by-step solution

Differentiate the function

Differentiate the function \(y=\frac{1}{x}+\sqrt{\cos x}\) using the rule \((\frac{1}{x})' = -\frac{1}{x^2}\) and \((\sqrt{\cos x})' = \frac{-\sin x}{2\sqrt{\cos x}}\). The derivative would be \(y' = -\frac{1}{x^2} - \frac{\sin x}{2\sqrt{\cos x}}\).

Evaluate the derivative at the given point

Substitute the point's \(x=\pi/2\) value into the derivative to get the value \(y'(x)= -\frac{1}{(\pi/2)^2}-\frac{\sin(\pi/2)}{2\sqrt{\cos(\pi/2)}} = -\frac{4}{\pi^2}\) as \(\cos(\pi/2) = 0\).

Verify the result using a graphing utility

After finding the derivative and its value at the given point, use a graphing utility. Plot the original function and its derivative and verify that the slope of the tangent at the given point on the original function matches the calculated derivative.

Key concepts

Derivative of a Function

Understanding the derivative of a function is fundamental in calculus. Imagine the derivative as a tool that provides the instantaneous rate of change of a function. Picture yourself driving a car—the speedometer shows your speed at every instant; that's what a derivative is like, but for functions instead of cars.

For a given function, denoted as \(y = f(x)\), the derivative, represented as \(f'(x)\) or \(\frac{dy}{dx}\), tells us how the function's output value changes as its input value changes. Taking the derivative of \(y = \frac{1}{x} + \sqrt{\cos x}\) involves applying rules for derivatives specific to each part of the function. For instance, the rule for the derivative of \(\frac{1}{x}\) indicates that the rate at which it decreases is \(-\frac{1}{x^2}\).

When finding derivatives, one may need to apply various rules like the power rule, product rule, quotient rule, or chain rule, depending on the complexity of the function.

Chain Rule in Calculus

The chain rule is a powerful derivative rule for handling the differentiation of composite functions. This occurs when you have a function nestled inside another function, like a Russian doll. In mathematical terms, if you have \(f(g(x))\), where both \(f\) and \(g\) are functions of \(x\), the chain rule dictates that the derivative is \(f'(g(x))g'(x)\).

Consider our initial problem where one term is \(\sqrt{\cos x}\). This is a composite function because the cosine function is inside the square root function. To differentiate, we apply the chain rule. First, take the outer function's derivative (the square root, which becomes \(\frac{1}{2\sqrt{\cos x}}\)) and multiply by the derivative of the inner function (the cosine, which becomes \(-\sin x\)). Combining them, you get \(\frac{-\sin x}{2\sqrt{\cos x}}\), the derivative of \(\sqrt{\cos x}\) with respect to \(x\).

Graphing Utility Verification

To confirm the calculations, we can utilize a graphing utility, an essential tool in modern calculus. By ploting the function \(y = \frac{1}{x} + \sqrt{\cos x}\) along with its derived function on a graph, we can gain a visual comprehension of the behavior of the function and its derivative.

In our problem, after finding our derivative analytically, we used a graphing utility to confirm the slope at the point \(\left(\frac{\pi}{2}, \frac{2}{\pi}\right)\). At this point on the graph of the original function, we expect the tangent line to have the same slope as the calculated derivative, \(-\frac{4}{\pi^2}\). By inspecting the graph, we should see that the tangent line's slope closely matches this value, thus serving as a visual and practical verification of our analytical work.

Trigonometric Functions Differentiation

Differentiating trigonometric functions is often required in calculus. Each trigonometric function has its own specific derivative. For example, the derivative of \(\sin x\) is \(\cos x\), while the derivative of \(\cos x\) is \(-\sin x\). These derivatives stem from the rate at which angles change with respect to the unit circle.

In our exercise, differentiating \(\sqrt{\cos x}\) required understanding the derivative of the cosine function. Here, the derivative was not simply \(-\sin x\), but needed to be adjusted because the cosine function was nested inside a square root. Remember that the derivatives of trigonometric functions must often be used in conjunction with other rules such as the chain rule to successfully differentiate more complex expressions involving trigonometry.