Question

In Exercises \(75-80\), evaluate the derivative of the function at the indicated point. Use a graphing utility to verify your result. \(\frac{\text { Function }}{y=\frac{1}{x}+\sqrt{\cos x}} \quad \frac{\text { Point }}{\left(\frac{\pi}{2}, \frac{2}{\pi}\right)}\)

Short answer

The derivative of the function \(y=\frac{1}{x}+\sqrt{\cos x}\) at the point \(\left(\frac{\pi}{2}, \frac{2}{\pi}\right)\) is undefined.

Step-by-step solution

Find the Derivative

Differentiate the function \(y=\frac{1}{x}+\sqrt{\cos x}\) using the power rule and the chain rule. The derivative of \(y=\frac{1}{x}\) is \(y'=-\frac{1}{x^{2}}\), and the derivative of \(\sqrt{\cos x}\) is \(\frac{1}{2\sqrt{\cos x}} \cdot -\sin x\). Therefore, the derivative of the entire function is \(y'=-\frac{1}{x^{2}} - \frac{\sin x}{2\sqrt{\cos x}}\).

Evaluate the Derivative at the Given Point

Substitute \(x=\frac{\pi}{2}\) into the derivative. Since \(\sin \frac{\pi}{2}=1\) and \(\cos \frac{\pi}{2}=0\), after substitution, we have \(y'=-\frac{1}{(\frac{\pi}{2})^{2}} - \frac{1}{2\sqrt{0}}\). The second part of this expression is undefined because it involves division by zero.

Conclusion

From the result in step 2, it's clear that the derivative at \(\left(\frac{\pi}{2}, \frac{2}{\pi}\right)\) is undefined because it involves division by zero. Therefore, the function \(y=\frac{1}{x}+\sqrt{\cos x}\) does not have a derivative at the point \(\left(\frac{\pi}{2}, \frac{2}{\pi}\right)\).

Key concepts

Chain Rule

The chain rule is a fundamental technique in calculus used for finding the derivative of composite functions. When we have a function that is composed of one function inside of another, the chain rule allows us to differentiate it step by step. For example, if we need to find the derivative of a function like \( \( h(x) = f(g(x)) \) \), we would apply the chain rule to get \( \ h'(x) = f'(g(x)) \cdot g'(x) \).

In the exercise, the chain rule is used to find the derivative of \( \( \sqrt{\cos x} \) \). Here, we effectively have two functions: the square root function and the cosine function. First, we differentiate the outer function, the square root, which gives us \( \( \frac{1}{2\sqrt{\cos x}} \) \), and then we multiply that by the derivative of the inner function, the cosine function, which is \( -\sin x \). Together, this application of the chain rule gives us the derivative \( \( \frac{-\sin x}{2\sqrt{\cos x}} \) \).

Understanding how to apply the chain rule is crucial for correctly evaluating and finding derivatives of various complex functions.

Power Rule

The power rule is another key concept in differential calculus, which states that the derivative of \( x^n \) with respect to \( x \) is \( nx^{n-1} \). It is applicable when dealing with polynomial functions or any terms where \( x \) is raised to a constant power.

In our original function \( y=\frac{1}{x} \), the term can be rewritten as \( x^{-1} \), which clearly lends itself to the power rule. Applying the power rule, we find the derivative to be \( -x^{-2} \), or \( -\frac{1}{x^2} \), and confirm that combining these rules allows us to find the derivatives of more complicated functions efficiently.

Graphing Utility Verification

Verifying the results of derivative calculations with graphic utilities can provide a visual confirmation that our solutions are correct. By plotting the function and its derivative, we can inspect critical points, such as where the derivative is zero or undefined, and understand the behavior of the function better.

For instance, in this exercise, after calculating the derivative, using a graphing utility to plot \( y \) and \( y' \) allows us to confirm that the derivative is indeed undefined at \( x=\frac{\pi}{2} \) because the graph will show a discontinuity or break at that point. This visual check supports the conclusion we reached through analytical methods.

Undefined Derivative

In calculus, an undefined derivative occurs at a point on a function where the slope of the tangent line cannot be determined. This can happen for various reasons, such as discontinuities in the function, corners, or vertical tangents. In the example of this exercise, the derivative becomes undefined due to the presence of the term \( \frac{1}{2\sqrt{\cos x}} \) where the cosine function equals zero. At \( x=\frac{\pi}{2} \) the denominator of this fraction would be zero, which is undefined in mathematics (since division by zero is not allowable).

Recognizing where a function's derivative might be undefined is important for understanding the overall behavior of the function, especially when it comes to analyzing limits, continuity, and differentiability of functions. Always consider the possibility of undefined derivatives when dealing with more complex functions and evaluating their derivatives at specific points.