Problem 54 - Calculus - AP Edition

Question

Find \(d^{2} y / d x^{2}\) $$y=x \tan \left(\frac{1}{x}\right)$$

Step-by-Step Solution

Answer

The second derivative is obtained by repeatedly applying the product and chain rules. Substitute and simplify to find \( \frac{d^2y}{dx^2} \).

Step 1: Differentiate Outer Function

Start by differentiating the given function using the product rule. Let the function be written as \[ y = u(x) v(x) \]where \( u(x) = x \) and \( v(x) = \tan\left(\frac{1}{x}\right) \). The derivative, \( \frac{dy}{dx} \), based on the product rule, is\[ \frac{dy}{dx} = u'(x) v(x) + u(x) v'(x) \].

Step 2: Differentiate Inner Functions

Now find the derivatives \( u'(x) \) and \( v'(x) \):1. \( u'(x) = 1 \) because the derivative of \( x \) is \( 1 \).2. To find \( v'(x) = \frac{d}{dx} \left( \tan\left(\frac{1}{x}\right) \right) \), apply the chain rule: \[ v'(x) = \sec^2\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) \].

Step 3: Substitute Derivatives Back

Substitute \( u'(x), v(x) \), and \( v'(x) \) into the product rule:\[ \frac{dy}{dx} = 1 \cdot \tan\left(\frac{1}{x}\right) + x \left( \sec^2\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) \right) \]which simplifies to:\[ \frac{dy}{dx} = \tan\left(\frac{1}{x}\right) - \frac{1}{x} \sec^2\left(\frac{1}{x}\right) \].

Step 4: Differentiate Again for the Second Derivative

Now differentiate the result to find \( \frac{d^2y}{dx^2} \):Start with: \[ \frac{d}{dx} \left( \tan\left(\frac{1}{x}\right) \right) - \frac{d}{dx} \left( \frac{1}{x} \sec^2\left(\frac{1}{x}\right) \right) \].We already have an expression for \( \frac{d}{dx} \left( \tan\left(\frac{1}{x}\right) \right) \) from before.For \( \frac{d}{dx} \left( \frac{1}{x} \sec^2\left(\frac{1}{x}\right) \right) \) use the product rule (where \( a(x) = \frac{1}{x} \) and \( b(x) = \sec^2\left(\frac{1}{x}\right) \)): \[ a'(x) b(x) + a(x) b'(x) \].

Step 5: Differentiate Inner Functions Again

Calculate:1. \( a'(x) = -\frac{1}{x^2} \).2. \( b'(x) = 2 \sec^2\left(\frac{1}{x}\right) \cdot \sec^2\left(\frac{1}{x}\right) \cdot -\frac{1}{x^2} = -\frac{2}{x^2} \sec^4\left(\frac{1}{x}\right) \).

Step 6: Substitute and Simplify

Substitute back into the product rule derivative for \( \frac{1}{x} \sec^2\left(\frac{1}{x}\right) \) and simplify:Finally, substitute this result and the derivative found for \( \tan\left(\frac{1}{x}\right) \) into the expression from Step 4 to find \( \frac{d^2y}{dx^2} \). Complete the simplification to find the second derivative in its simplest form.

Key Concepts

Product Rule

The product rule is a fundamental differentiation technique used when you want to differentiate the product of two functions. If you have a function that is composed of two separate functions multiplied together, such as \( y = u(x) \times v(x) \), the product rule states that the derivative of this product is:

• \( \frac{dy}{dx} = u'(x) \times v(x) + u(x) \times v'(x) \)

This means you differentiate the first function, multiply it with the second function, and then add it to the product of the first function and the derivative of the second function.
In the given exercise, \( u(x) = x \) and \( v(x) = \tan\left(\frac{1}{x}\right) \). To find \( \frac{dy}{dx} \), you first differentiate \( u(x) \) giving \( u'(x) = 1 \), and use the chain rule to find \( v'(x) \). This helps you manage complex expressions by breaking them into simpler parts, ensuring accuracy and clarity in finding derivatives of products.

Chain Rule

The chain rule is another vital differentiation technique, especially for functions composed of nested functions, or where one function is inside another. The rule allows you to differentiate such composite functions by focusing on each layer separately.
The chain rule formula is:

• \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \)

Applying the chain rule means that you first differentiate the outer function while keeping the inner function unchanged, and then multiply by the derivative of the inner function.
In the original exercise, the chain rule is employed when differentiating \( v(x) = \tan\left(\frac{1}{x}\right) \). Here, the outer function is \( \tan(u) \) (where \( u = \frac{1}{x} \)), and its derivative is \( \sec^2(u) \). The inner function, \( \frac{1}{x} \), is then differentiated to give \( -\frac{1}{x^2} \). The combination of these two derivatives gives the full \( v'(x) \). This method is essential for working with complex composite functions and ensures that each component is treated properly.

Differentiation Techniques

Differentiation is a powerful tool in calculus used to find how a function changes at any point. Different rules and techniques allow you to handle various types of functions and find their derivatives accurately.
Apart from the basic derivative rules, here are a few important differentiation techniques that are often used:

• Power Rule: For any function \( f(x) = x^n \), the derivative is \( nx^{n-1} \).
• Quotient Rule: Used when you have one function divided by another: \( \frac{d}{dx} \left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \).
• Trigonometric Derivatives: Involves derivatives of sine, cosine, tangent, and other trig functions.

These techniques, like the product and chain rules, are essential for tackling functions that involve more than just one operation, such as products or compositions of functions.
The exercise we solved involved both the product rule and the chain rule, showcasing how multiple techniques can be combined to solve complex differentiation problems. Understanding each differentiation technique helps you apply them correctly and find derivatives efficiently.