91Ó°ÊÓ

The chain rule is a fundamental technique used in calculus for differentiating compositions of functions. It is essential for solving complex derivative problems where one function is nested inside another. Imagine you have two functions, \( f(x) \) and \( g(x) \), where \( f(x) \) is the outer function and \( g(x) \) is the inner function. The chain rule states that the derivative of their composition \( (f \circ g)(x) \) is \( f'(g(x)) \cdot g'(x) \).

In the context of verifying solutions to differential equations, it helps us to differentiate terms like \( \sin(t)\ln(\cos(t)) \) where \( \ln(\cos(t)) \) is the inner function and \( \sin(t) \) could be interpreted as an outer function. Applying the chain rule, one would take the derivative of the outer function and multiply it by the derivative of the inner function, successfully breaking down complex derivatives into simpler parts.

The product rule is another indispensable tool used to differentiate expressions where two functions are multiplied by each other. Mathematically, if you have two functions \( u(t) \) and \( v(t)\), their derivative \( u'(t)v(t) \) plus \( u(t)v'(t) \) gives you the derivative of the product \( u(t)v(t) \). This rule is necessary when handling terms such as \( t\sin(t) \) where 't' and \( \sin(t) \) are functions of 't' being multiplied.

During the verification of a differential equation solution, you will often need to apply the product rule to correctly differentiate these products. This step is crucial in obtaining accurate expressions for the first and subsequent derivatives.

Calculating the first derivative of a function is equivalent to finding the rate at which the function's value changes with respect to its variable. It's the cornerstone of differential equations as it represents the instantaneous rate of change. For instance, given a function \( y = f(t) \), the first derivative \( y' = f'(t) \) tells us how \( y \) changes as \( t \) changes. In differential equations like the one featured in our exercise, calculating the first derivative of \( y \) is typically the starting point to verify if a provided function is indeed a solution.

When we calculated \( y' \) for our exercise by utilizing both the chain rule and the product rule, we got \( y' = t\cos(t) - \sin(t)\ln(\cos(t)) \), which simplified the expression and prepared us for the next differentiation step.

The second derivative of a function \( f(t) \) gives us the curvature or the acceleration of the function's graph. In our case, the second derivative \( y'' \) shows how the rate of change of \( y' \) itself changes. For the verification process of a differential equation, we compute the second derivative to evaluate how the solution's curvature conforms to the differential equation.

In our exercise, calculating \( y'' \) involves differentiating \( y' = t\cos(t) - \sin(t)\ln(\cos(t)) \) to arrive at an expression that includes the original function \( y \) and \( y' \) to check against \( \sec(t) \) as given in the differential equation. This step confirms whether the function behaves according to the dynamics dictated by the differential equation over the given interval.