91Ó°ÊÓ

Differentiation is a fundamental concept in calculus. It involves finding the rate at which a function is changing at any given point. The derivative of a function indicates the slope of the tangent line to the function's graph at that point.

The process of differentiation can be broken down into several rules and steps. These rules include the power rule, product rule, quotient rule, and chain rule (which we'll cover later). Each rule helps make the process of finding derivatives more straightforward.

In our example, we start by differentiating the function \(s(t) = \cos(at + b)\). To do this, we need to apply the chain rule.

The chain rule is crucial when dealing with composite functions, where one function is nested inside another. Essentially, it helps you differentiate a function with respect to an inner function.

The chain rule states that if you have a function \(h(x) = f(g(x))\), then the derivative \(h'(x)\) is found by multiplying the derivative of the outer function \(f\) evaluated at \(g(x)\) by the derivative of the inner function \(g(x)\). In notation form:

• \( \frac{d}{dx} [f(g(x))] = f'(g(x)) \times g'(x) \)

In the exercise, we need to find the first derivative of \(s(t) = \cos(at + b)\). Here:

• Outer function: \( \cos(x) \) with derivative \( -\sin(x) \)
• Inner function: \( at + b \) with derivative \( a \)

So, the first derivative is:

• \( s'(t) = -\sin(at + b) \times a = -a\sin(at + b)\)

Then, to find the second derivative, we differentiate \( s'(t) \) again, applying the chain rule once more:

\( s''(t) = -a\sin(at + b)' = -a(\cos(at + b) \times a) = -a^2\cos(at + b) \)

Trigonometric functions like sine and cosine are fundamental in calculus and require special rules for differentiation and integration. These functions are periodic and show how angles relate to ratios in right triangles.

Consider the function \( \cos(x) \). Its derivative is \( -\sin(x) \). This derivative represents the rate of change of the cosine function.

When differentiating trigonometric functions like in our example, you often use their well-known derivatives:

• Derivative of \( \sin(x) \) is \( \cos(x) \)
• Derivative of \( \cos(x) \) is \( -\sin(x) \)

Thus, when we differentiate \(s(t) = \cos(at + b)\), we apply these properties and the chain rule. After finding the first derivative as \( s'(t) = -a\sin(at + b)\), we obtain the second derivative:

\( s''(t) = -a^2\cos(at + b) \)

This shows how trigonometric differentiation and the chain rule work together to solve calculus problems.