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An initial-value problem is a type of differential equation coupled with a given initial condition. The goal is to find a function that satisfies both the differential equation and the initial condition. This essentially means that

• the function must meet the specified differential equation’s requirements
• at a particular point, usually denoted by \( x = 0 \), the function must equal the given initial value

In our example, we looked at the initial-value problem where we needed to confirm the function \( y = \sin x \cos x - \cos x \) is a solution for \[ y' + (\tan x)y = \cos^2 x \quad \text{with}\, y(0) = -1 \]This involves ensuring that when \( x = 0 \), the value of \( y \) is \(-1\). By substituting \( x = 0 \) into \( y \), we verified the equation held true. Understanding this helps predict how a function behaves based on initial circumstances.

Derivatives are fundamental in calculus as they describe how a function changes. The derivative of a function \( y \), symbolized as \( y' \) or \( \frac{dy}{dx} \), represents its rate of change or slope.
When solving the given differential equation, we first computed the derivative of \( y = \sin x \cos x - \cos x \). In calculus, we focus on understanding the behavior of a function at a microscale, i.e., at an infinitesimally small level.
For our specific function, we had:

• The derivative of \( \sin x \cos x \) became \( \cos^2 x - \sin^2 x \)
• The derivative of \( -\cos x \) became \( \sin x \)

Thus, we concluded that the overall derivative \( y' \) is:\[ y' = \cos^2 x - \sin^2 x + \sin x \]
Knowing this derivative allowed us to substitute back into and verify the original differential equation.

The product rule is a basic tool in calculus for finding the derivative of a product of two functions. This rule is especially useful because many complex functions are compositions of simpler functions. The formula for the product rule is:\[ (uv)' = u'v + uv' \]where \( u \) and \( v \) are functions of \( x \). In other words, you differentiate the first function and multiply it with the second function, then do the reverse and sum them up.
In solving our problem, the derivative of \( \sin x \cos x \) using the product rule was key. In the solution process, \( u = \sin x \) and \( v = \cos x \), hence:

• \( u' = \cos x \)
• \( v' = -\sin x \)
• So, \( (\sin x \cos x)' = (\sin x)(-\sin x) + (\cos x)(\cos x) = -\sin^2 x + \cos^2 x \)

This demonstrates the product rule at work, showing the importance of knowing how to tackle derivatives of function products.

Trigonometric identities are equations that involve trigonometric functions that are true for every value of the variables involved. They play an essential role in simplifying expressions and solving equations. These identities make complex calculations much manageable.
In the given problem, we used specific trigonometric identities:

• For substitution into the equation: \( \tan x = \frac{\sin x}{\cos x} \)
• We also encountered identities like \( \cos^2 x + \sin^2 x = 1 \) which, although not directly used, help in many similar problems

These identities allowed us to check the simplifications properly when substituting back into the differential equation, ensuring everything balanced on either side. Mastery of these trigonometric identities is vital for anyone delving deep into calculus.