Problem 3 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find the first and second derivatives of the following. (a) \(f(x)=5 \cos x\) (b) \(g(x)=-3 \sin (2 x)\) (c) \(h(x)=0.5 \tan x\) (d) \(j(x)=2 \sin x \cos x\)

Step-by-Step Solution

Answer

First and second derivatives of the functions are: (a) \(f'(x)=-5\sin x\) and \(f''(x)=-5\cos x\)(b) \(g'(x)=-6\cos(2x)\) and \(g''(x)=12\sin(2x)\)(c) \(h'(x)=0.5\sec^2x\) and \(h''(x)=\sec x \tan x\)(d) \(j'(x)=2(\cos^2x-\sin^2x)\) and \(j''(x)=-4\sin x \cos x\)

Step 1: Differentiation of f(x)

Differentiate \(f(x)=5\cos x\) using the rule that the derivative of \(\cos x\) is \(-\sin x\). This gives the first derivative as \(f'(x)=-5\sin x\). Then differentiate \(f'(x)=-5\sin x\) to get the second derivative. The derivative of \(\sin x\) is \(\cos x\), so the second derivative is \(f''(x)=-5\cos x\)

Step 2: Differentiation of g(x)

Differentiate \(g(x)=-3\sin(2x)\) by applying the chain rule, which states that the derivative of \(f(g(x))\) is \(f'(g(x)) \cdot g'(x)\). Here, \(f(u) = -3\sin u\) and \(u=2x\). The derivative of \(-3\sin u\) is \(-3\cos u\) and the derivative of \(2x\) is 2. Using the chain rule gives the first derivative as \(g'(x)=-6\cos(2x)\). Differentiate \(g'(x)=-6\cos(2x)\) to get the second derivative, applying the chain rule again gives \(g''(x)=12\sin(2x)\)

Step 3: Differentiation of h(x)

Differentiate \(h(x)=0.5\tan x\) using the rule that the derivative of \(\tan x\) is \(\sec^2x\). The first derivative is \(h'(x)=0.5\sec^2x\). Then, differentiate \(h'(x)=0.5\sec^2x\) to get the second derivative. The derivative of \(\sec^2x\) is \(2\sec x \tan x\), so the second derivative is \(h''(x)=0.5(2\sec x \tan x) = \sec x \tan x\)

Step 4: Differentiation of j(x)

Differentiate \(j(x)=2\sin x\cos x\) using the product rule, which states that the derivative of \(f(x)g(x)\) is \(f'(x)g(x) + f(x)g'(x)\). Applying this rule gives the first derivative as \(j'(x)=2(\cos^2x-\sin^2x)\). Differentiating \(j'(x)=2(\cos^2x-\sin^2x)\) to get the second derivative using the chain rule and the rules for derivatives of \(\sin x\) and \(\cos x\) yields \(j''(x)=-4\sin x \cos x\)

Key Concepts

Chain Rule

When you’re dealing with a function composed of other functions, the chain rule is your indispensable tool. Imagine you have a function like \(-3 \sin(2x)\). Here, we have an outer function, \(-3 \sin(u)\), and an inner function, \(u = 2x\). The chain rule helps us differentiate such compositions.

The chain rule states that the derivative of a composite function \(f(g(x))\) is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In formal terms this is expressed as:

• \(f'(g(x)) \cdot g'(x)\)

In our example, differentiating \(-3 \sin(u)\) gives \(-3 \cos(u)\), and differentiating \(2x\) gives \(2\). Putting it all together using the chain rule, the first derivative is \(-6 \cos(2x)\).

Understanding the chain rule is crucial because it appears so frequently in various calculus problems. By breaking down a difficult problem into simpler parts, the chain rule makes even the most complex functions solvable.

Product Rule

The product rule is essential when you have two functions being multiplied together. For example, consider \(j(x) = 2 \sin x \cos x\). In this situation, we use the product rule to find the derivative.

The rule can be memorized with the formula for two functions \(f(x)\) and \(g(x)\), given by:

• \(f(x)g(x) \rightarrow f'(x)g(x) + f(x)g'(x)\)

For our function, let's assign \(f(x) = 2 \sin x\) and \(g(x) = \cos x\). Then the derivatives become \(f'(x) = 2 \cos x\) and \(g'(x) = -\sin x\). Applying the product rule gives us the derivative: \(j'(x) = 2(\cos^2x - \sin^2x)\).

Knowing how to apply the product rule effectively allows you to tackle a wide range of problems where functions are multiplied. It simplifies the derivative process and is often used alongside the chain rule and other function rules.

Trigonometric Functions

Trigonometric functions like \(\sin x\), \(\cos x\), and \(\tan x\) frequently appear in calculus, and understanding their derivatives is key to mastering them. Let's walk through some examples you might encounter:

- **Derivative of \(\sin x\):** The derivative is \(\cos x\). - **Derivative of \(\cos x\):** The derivative is \(-\sin x\).- **Derivative of \(\tan x\):** The derivative is \(\sec^2 x\).

Differentiating these trigonometric functions allows you to solve complex problems involving oscillations and waves. For instance, if you have \(f(x) = 5 \cos x\), differentiating using the known derivative of \(\cos x\) gives \(-5 \sin x\) as the derivative. Doing it again yields the second derivative: \(-5 \cos x\).

The beauty of calculus with trigonometric functions rises from its pattern recognition; the sine and cosine functions cycle between each other upon differentiation, and understanding this pattern provides a deeper insight into the behavior of these functions.