Problem 194 - Calculus Volume 1

Question

Find \(\frac{d^{2} y}{d x^{2}}\) for the given functions. $$ y=\frac{1}{x}+\tan x $$

Step-by-Step Solution

Answer

\(\frac{d^{2} y}{d x^{2}} = \frac{2}{x^{3}} + 2\sec^{2} x \tan x\).

Step 1: Understand the Problem Statement

The problem asks for the second derivative of the function \(y = \frac{1}{x} + \tan x\). This means we need to find the derivative of this function twice with respect to \(x\).

Step 2: Differentiate the First Term \(\frac{1}{x}\)

Calculate the first derivative of \(\frac{1}{x}\). We can use the power rule by rewriting \(\frac{1}{x}\) as \(x^{-1}\). Now differentiate: \(\frac{d}{dx} x^{-1} = -x^{-2} = -\frac{1}{x^{2}}\).

Step 3: Differentiate the Second Term \(\tan x\)

The derivative of \(\tan x\) is well-known: \(\frac{d}{dx} \tan x = \sec^{2} x\).

Step 4: Combine First Derivatives

Combine the first derivatives obtained from Steps 2 and 3. The first derivative of the function is \(\frac{dy}{dx} = -\frac{1}{x^{2}} + \sec^{2} x\).

Step 5: Differentiate \(-\frac{1}{x^{2}}\) Again

Find the second derivative of \(-\frac{1}{x^{2}}\) as follows: Consider \(-x^{-2}\), and differentiate: \(\frac{d}{dx} (-x^{-2}) = 2x^{-3} = \frac{2}{x^{3}}\).

Step 6: Differentiate \(\sec^{2} x\) Again

The second derivative of \(\sec^{2} x\) involves the chain rule and is calculated as follows: \(\frac{d}{dx} \sec^{2} x = \frac{d}{dx} ((\sec x)^{2}) = 2\sec x \cdot \sec x \tan x = 2\sec^{2} x \tan x\).

Step 7: Combine Second Derivatives

The second derivative of the function is \(\frac{d^{2} y}{d x^{2}} = \frac{2}{x^{3}} + 2\sec^{2} x \tan x\).

Key Concepts

Power Rule

The Power Rule is a fundamental tool in calculus for finding derivatives. If you have an expression in the form of \(x^n\), the derivative is found using the formula \(\frac{d}{dx} x^n = nx^{n-1}\). This means you bring the exponent down as a coefficient and then decrease the exponent by one.

Let's apply this to the problem where we need to differentiate \(\frac{1}{x}\). First, it's helpful to rewrite \(\frac{1}{x}\) as \(x^{-1}\). Now, applying the Power Rule, we differentiate:\

• The exponent \(n = -1\) comes down in front of \(x\) as a coefficient: \(-1 \cdot x^{(-1-1)} = -x^{-2}\).
• This simplifies to \(-\frac{1}{x^2}\).

This technique makes dealing with negative exponents straightforward and is indispensable when working through calculus problems involving powers.

Chain Rule

The Chain Rule is a method for finding the derivative of composite functions. It's essential when you have a function nested inside another function. The rule states: if you have a function \(h(x) = f(g(x))\), then the derivative \(h'(x)\) is \(f'(g(x)) \cdot g'(x)\). In simple terms, differentiate the outer function, and multiply it by the derivative of the inner function.

In the solution, we see the Chain Rule in action when differentiating \(\sec^2 x\), which involves an inner and outer function. Here, the outer function is \((\sec x)^2\), and the inner function is \(\sec x\). The derivative of \((\sec x)^2\) is found by

• First finding \(2 \cdot \sec x\) by applying the power and bringing down the 2.
• Multiplying by the derivative of \(\sec x\), which is \(\sec x \tan x\).

Therefore, the derivative is \(2 \cdot (\sec x) \cdot (\sec x \tan x) = 2 \sec^2 x \tan x\). This is a practical example of how the Chain Rule works to differentiate nested trigonometric functions in calculus.

Trigonometric Derivatives

Derivatives of trigonometric functions are some of the most commonly used in calculus. Each trigonometric function has a specific derivative that helps us understand how these functions change with respect to a variable like \(x\).

In this exercise, we encounter the derivative of \(\tan x\) and \(\sec^2 x\). Remember these key derivatives:

• The derivative of \(\tan x\) is \(\sec^2 x\).
• The derivative of \(\sec x\) involves the product of \(\sec x\) and \(\tan x\), specifically \(\sec x \tan x\).

By familiarizing ourselves with these derivatives, solving problems that involve trigonometric functions becomes much easier. In our problem, once you calculate \(\frac{d}{dx} \tan x = \sec^2 x\), combining these derivatives gives us a clearer picture of how the function \(y = \frac{1}{x} + \tan x\) behaves when differentiated. Understanding these basic trigonometric derivatives allows us to tackle more complex calculus problems with confidence.