Problem 56 - Thomas Calculus

Question

In Exercises \(53-58,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x\) . $$ f(u)=u+\frac{1}{\cos ^{2} u}, \quad u=g(x)=\pi x, \quad x=1 / 4 $$

Step-by-Step Solution

Answer

The value is \(5\pi\).

Step 1: Understand the Problem

We need to evaluate the derivative of the composite function \( (f \circ g)(x) \) at \( x = \frac{1}{4} \). The functions given are \( f(u) = u + \frac{1}{\cos^2 u} \) and \( u = g(x) = \pi x \).

Step 2: Calculate g(x) and its Derivative

The function \( g(x) = \pi x \). Therefore, at \( x = \frac{1}{4} \), \( g(x) = \pi \left(\frac{1}{4}\right) = \frac{\pi}{4} \). The derivative of \( g(x) \) with respect to \( x \) is \( g'(x) = \pi \).

Step 3: Calculate f(u) and its Derivative

We have \( f(u) = u + \frac{1}{\cos^2 u} \). The derivative of \( f \) with respect to \( u \) is found using the chain rule: \( f'(u) = 1 + \frac{d}{du} \left( \frac{1}{\cos^2 u} \right) \).Using the chain rule, \( \frac{d}{du} \left( \frac{1}{\cos^2 u} \right) = \frac{2\sin u}{\cos^3 u} \), so \( f'(u) = 1 + \frac{2 \sin u}{\cos^3 u} \).

Step 4: Apply the Chain Rule for Derivatives

The derivative of the composite function \( (f \circ g)^{\prime}(x) \) is given by \( f'(g(x)) \cdot g'(x) \).Compute \( f'(\frac{\pi}{4}) \):- \( \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \) and \( \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \).- \( f'(\frac{\pi}{4}) = 1 + \frac{2 \left(\frac{\sqrt{2}}{2}\right)}{\left(\frac{\sqrt{2}}{2}\right)^3} = 1 + 4 = 5 \).Therefore, \( (f \circ g)^{\prime}(\frac{1}{4}) = 5 \cdot \pi = 5\pi \).

Step 5: Final Answer

The value of the derivative \( (f \circ g)^{\prime}(x) \) at \( x = \frac{1}{4} \) is \( 5\pi \).

Key Concepts

Composite Functions

When dealing with composite functions, we're essentially looking at a combination of two functions, where one function is applied to the result of another. In simpler terms, think of it as layering one function on top of another.
For example, if you have two functions, say \( f(u) \) and \( g(x) \), the composite function \( (f \circ g)(x) \) involves first applying \( g(x) \) and then feeding that result into \( f(u) \).
A common example is if \( g(x) \) represents an angle in radians, like \( \pi x \), and \( f(u) \) includes trigonometric identities relying on this angle. The exercise deals precisely with this kind of scenario, paving the way for the application of calculus techniques such as derivatives.

Derivatives of Trigonometric Functions

Trigonometric functions, such as \( \cos(u) \) and \( \sin(u) \), are foundational in calculus, and knowing how to differentiate them is crucial.
The derivative of \( \sin(u) \) with respect to \( u \) is \( \cos(u) \), and the derivative of \( \cos(u) \) is \(-\sin(u) \). But when functions involve more complexity, like \( \frac{1}{\cos^2 u} \), you need to utilize the chain rule to successfully differentiate.
The exercise showcases how the chain rule is used to calculate the derivative \( \frac{d}{du} \left( \frac{1}{\cos^2 u} \right) = \frac{2 \sin u}{\cos^3 u} \). It requires using the power rule paired with understanding the trigonometric function's derivative, simplifying it by applying basic trigonometric identities.

Calculus Exercises

Calculus exercises often involve applying the chain rule, especially when handling composite functions. The chain rule is a fundamental principle used to take the derivative of compositions of functions. It states that the derivative of \( (f \circ g)(x) \) is \( f'(g(x)) \cdot g'(x) \).
In the exercise, \( g(x) = \pi x \) with its derivative being \( g'(x) = \pi \), and \( f(u) = u + \frac{1}{\cos^2 u} \) with the derivative \( f'(u) = 1 + \frac{2 \sin u}{\cos^3 u} \). By plugging in \( u = \frac{\pi}{4} \), you evaluate \( f' \) and find \( f'(\frac{\pi}{4}) = 5 \).

• Ultimately, finding \( (f \circ g)'(\frac{1}{4}) = f'(g(\frac{1}{4})) \cdot g'(\frac{1}{4}) = 5 \cdot \pi = 5\pi \) brings you to the solution.

Applying these principles can be challenging at first, but breaking them down into steps, as demonstrated in the exercise, can make calculus less daunting and more approachable.