Question

Finding a Second Derivative In Exercises \(85-90\) , find the second derivative of the function. $$ f(x)=\sin x^{2} $$

Short answer

The second derivative of the function \( f(x) = \sin(x^{2}) \) is \( -4x^{2}\sin(x^{2}) + 2\cos(x^{2}) \)

Step-by-step solution

Find the First Derivative

Calculating the first derivative using chain rule. The derivative of \( \sin(x) \) is \( \cos(x) \) and the derivative of \( x^{2} \) is \( 2x \). Using the chain rule, the derivative of \( \sin(x^{2}) \) is \( \cos(x^{2}) \cdot 2x \) which simplifies to \( 2x \cos(x^{2}) \).

Find the Second Derivative

Differentiating again, the derivative of \( 2x \cos(x^{2}) \) can be found by using the product rule. The derivative of \( 2x \) is \( 2 \) and the derivative of \( \cos(x^{2}) \) is \( -\sin(x^{2}) \cdot 2x \), again using the chain rule. So, the second derivative of \( f(x) \) is \( 2(-\sin(x^{2}) \cdot 2x) + \cos(x^{2}) \cdot 2 \), which simplifies to \( -4x^{2}\sin(x^{2}) + 2\cos(x^{2})\).

Key concepts

Chain Rule

The chain rule is a principle in calculus that provides a method for finding the derivative of composite functions. Simply put, if you have a function that's made up of other functions, the chain rule can help you find its derivative.

For instance, if you have a function like \( f(x) = \text{sin}(x^2) \), the inner function is \( g(x) = x^2 \) and the outer function is \( h(g) = \text{sin}(g) \). When deriving \( f(x) \) with respect to \( x \), you first differentiate the outer function with respect to the inner function (so you get \( \text{cos}(g) \) when \( g = x^2 \)) and then multiply it by the derivative of the inner function. Mathematically, the formula is written as \( \frac{df}{dx} = \frac{dh}{dg} \times \frac{dg}{dx} \), which in our example translates to \( \frac{df}{dx} = \text{cos}(x^2) \times 2x \).

This step-by-step progression elegantly demonstrates the chain rule in action, allowing the complex derivative to be broken down into simpler parts that are easier to compute individually.

Product Rule

The product rule is another fundamental theorem in calculus, vital for differentiating expressions where two functions are multiplied by each other. Traditional algebraic methods don't work in this case, and thus the product rule is essential.

The rule states that to differentiate a product of two functions \( u(x) \) and \( v(x) \), you take the derivative of the first function and multiply it by the second function, and then add the product of the first function and the derivative of the second function. The formula is expressed as \( \frac{d}{dx}(u \times v) = u' \times v + u \times v' \).

Applying this to the problem at hand, if we consider \( u(x) = 2x \) and \( v(x) = \text{cos}(x^2) \), the first derivative of \( u(x) \) is 2, and applying the chain rule for \( v(x) \), we get \( v'(x) = -\text{sin}(x^2) \times 2x \). Piecing it together, the second derivative of \( f(x) \) becomes \( 2 \times -\text{sin}(x^2) \times 2x + 2x \times 2 \), simplifying to \( -4x^2\text{sin}(x^2) + 4 \times \text{cos}(x^2) \), which corrects the previous simplification error.

Derivatives of Trigonometric Functions

Understanding the derivatives of trigonometric functions is crucial when dealing with calculus problems involving sine, cosine, tangent, and other trigonometric functions. Each function has a specific derivative that is used when applying the chain rule or product rule.

For instance, the derivative of \( \text{sin}(x) \) is \( \text{cos}(x) \), while the derivative of \( \text{cos}(x) \) is \( -\text{sin}(x) \). Conversely, the derivatives can become more complex when the trigonometric functions involve composite functions, like \( \text{sin}(x^2) \), which is where the chain rule comes into play.

In our problem, the challenge was to find the derivative of \( f(x) = \text{sin}(x^2) \). Alignment with the rule for the derivative of \( \text{sin} \), as well as the subsequent differentiation of \( \text{cos} \), is vital for finding accurate results. Mastering the derivatives of basic trigonometric functions simplifies the process of finding derivatives of more complex trigonometric expressions.