Problem 15 - Calculus - AP Edition

Question

Find \(f^{\prime}(x)\) $$f(x)=\sin \left(\frac{1}{x^{2}}\right)$$

Step-by-Step Solution

Answer

The derivative is \( f'(x) = -\frac{2 \cos\left( \frac{1}{x^2} \right)}{x^3} \).

Step 1: Understand the Function Structure

The function we need to differentiate is given by the expression \( f(x) = \sin\left( \frac{1}{x^2} \right) \). This is a composition of functions, where we have an outer function, sine, and an inner function, \( \frac{1}{x^2} \).

Step 2: Use the Chain Rule

Since the structure involves a composition of functions, we will use the chain rule for differentiation. The chain rule states that if you have a function \( g(h(x)) \), the derivative \( g'(x) \) is \( g'(h(x)) \cdot h'(x) \).

Step 3: Differentiate the Outer Function

First, differentiate the outer function where the input is \( \frac{1}{x^2} \). The derivative of \( \sin(u) \) with respect to \( u \) is \( \cos(u) \). So, the derivative with respect to the inner function \( \frac{1}{x^2} \) is \( \cos\left( \frac{1}{x^2} \right) \).

Step 4: Differentiate the Inner Function

The inner function is \( \frac{1}{x^2} = x^{-2} \). Its derivative with respect to \( x \) using the power rule \( \frac{d}{dx} [x^n] = nx^{n-1} \) is \( -2x^{-3} = -\frac{2}{x^3} \).

Step 5: Apply the Chain Rule

Substitute the derivatives found in Steps 3 and 4 into the chain rule formula: \[ f'(x) = \cos\left( \frac{1}{x^2} \right) \cdot \left( -\frac{2}{x^3} \right) \]. This simplifies to the final derivative: \[ f'(x) = -\frac{2 \cos\left( \frac{1}{x^2} \right)}{x^3} \].

Key Concepts

Function Composition

Understanding function composition is crucial in calculus, particularly when dealing with complex expressions. A composite function involves one function inside another function, similar to nesting. In the given exercise, we have a function \( f(x) = \sin\left( \frac{1}{x^2} \right) \). Here:

• The outer function is the sine function \( \sin(u) \).
• The inner function is \( \frac{1}{x^2} \).

Function composition makes solving such functions more structured. We differentiate by addressing each function separately, starting with the outer function while keeping the inner function unchanged. This orderly approach helps simplify otherwise complex problems.

Power Rule

The power rule is a fundamental tool in calculus used to differentiate expressions in the form \( x^n \). It states that the derivative of \( x^n \) is \( nx^{n-1} \). In our exercise, the inner function can be expressed as \( x^{-2} \). Using the power rule, the derivative is:

• Multiply the exponent by the coefficient: \( -2 \times 1 = -2 \)
• Reduce the exponent by one: \( -2 - 1 = -3 \)

Thus, the derivative of the inner function \( x^{-2} \) is \( -2x^{-3} \), or \( -\frac{2}{x^3} \). This step provides the necessary component to use in the chain rule when addressing function composition.

Trigonometric Functions

Trigonometric functions, such as sine, cosine, and tangent, appear often in calculus problems, alongside their derivatives. The sine function \( \sin(u) \) has a straightforward derivative: \( \cos(u) \). In the context of function composition:

• You maintain the general form \( \cos(u) \) even if \( u \) is itself a complex expression.
• For our exercise, the derivative of the outer function \( \sin(u) \) where \( u = \frac{1}{x^2} \) is \( \cos\left( \frac{1}{x^2} \right) \).

Trigonometric derivatives keep the computations consistent. When used in conjunction with other rules, like the power and chain rules, they help achieve the desired solution efficiently. Understanding these derivatives assists in tackling broader calculus problems.