Question

Finding a Second Derivative In Exercises \(91-98\) , find the second derivative of the function. $$ f(x)=x \sin x $$

Short answer

The second derivative of the function \(f(x) = xsin(x)\) is \(f''(x) = cos(x) - x sin(x)\).

Step-by-step solution

Identify the Functions

Identify the two functions that are being multiplied: \(f(x) = x \cdot sin(x)\). Let \(u = x\), and \(v = sin(x)\).

Differentiate

We use the differentiation product rule which states that \( (uv)' = u'v + uv'\) where \(u' \) and \(v'\) are the derivatives of \(u\) and \(v\) respectively. Differentiating \(u\) and \(v\) gives \(u' = 1\) and \(v' = cos(x)\). Apply these into the product rule to get the first derivative, \(f'(x) = 1 \cdot sin(x) + x \cdot cos(x)\).

Find the Second Derivative

Next, find the second derivative, \(f''(x)\), by differentiating \(f'(x)\) using product rule again. This gives two new functions: \(u = x\), \(v = cos(x)\) with derivatives \(u' = 1\), and \(v' = -sin(x)\). The second derivative is thus given by \(f''(x) = 1 \cdot cos(x) + x \cdot -sin(x) + sin(x)\). Simplifying gives \(f''(x) = cos(x) - x sin(x)\).

Key concepts

Product Rule Differentiation

The product rule is a fundamental technique of differentiation that comes into play when you are dealing with the multiplication of two functions. Imagine you are trying to find the rate at which a product of two quantities changes; this is where the product rule shines.

In calculus, the product rule is formally stated as if you have two functions, let's call them \( u(x) \) and \( v(x) \), their derivative when multiplied together (\( u(x) \cdot v(x) \) is not simply the product of their individual derivatives. The correct derivative, using the product rule, is \( (u \cdot v)' = u' \cdot v + u \cdot v' \) where \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \) respectively.

This means you take the derivative of the first function and multiply it by the second function as is, and then you add the product of the first function times the derivative of the second function. It's like a dance between differentiation and multiplication ensuring that all parts of the functions are accounted for in the rate of change.

To smoothly apply the product rule, keep in mind:

• Always identify the two functions being multiplied.
• Differentiate each function separately before applying the rule.
• Apply the derivatives correctly according to the rule without mixing them up.

Remember that the product rule is essential when dealing with any function that is the multiplication of two or more functions, especially when they cannot be simplified into a single term before differentiation.

First Derivative

The first derivative of a function represents the rate at which the function's value changes with respect to change in the input value. In other words, it's concerned with the slope of the tangent line to the function at any point. Calculating the first derivative is the cornerstone of finding how fast or slow things are changing in calculus.

For a function \( f(x) \), the first derivative is denoted \( f'(x) \) or \( \frac{df}{dx} \) and it gives us vital information about the behavior of \( f(x) \). For example, it can tell us about the function's increasing or decreasing nature, and also about relative maximums and minimums.

To calculate the first derivative:

• You start by applying the relevant differentiation rules based on the function's structure, like the product rule, quotient rule, or chain rule.
• Simplify the derivative as much as possible for an easier interpretation of results.

It's crucial to be precise with the application of differentiation rules; a small error can lead to incorrect conclusions about a function's behavior. Hence, practice and a keen eye for detail are essential to mastering the calculation of the first derivative.

Trigonometric Functions Differentiation

Differentiating trigonometric functions is an integral part of calculus, as these functions model various phenomena in physics and engineering. The most common trigonometric functions in calculus are sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)), each with its own specific rate of change.

When you differentiate \( \sin(x) \) the result is \( \cos(x) \) and differentiating \( \cos(x) \) gives you \( -\sin(x) \)—note the negative sign which is due to the direction of \( \cos \)’s slope as \( x \) increases. Differentiating \( \tan(x) \) gets a bit more complex, yielding \( \sec^2(x) \), where \( \sec(x) \) is the secant function.

• The derivatives of trigonometric functions reveal the cyclical nature of their rate of change.
• It is crucial to remember these basic rules of differentiation for trigonometric functions to simplify more complex differentiation problems.
• For higher trigonometric functions like \( \sec(x) \) or \( \csc(x) \), their derivatives involve other trigonometric functions and often their squares.

Understanding the derivatives of trigonometric functions is not only about memorizing these rules but also about recognizing their patterns and how they relate to the unit circle, a foundational concept in trigonometry that elegantly illustrates why these derivatives make sense.