91Ó°ÊÓ

When differentiating a function that is a quotient of two other functions, like \( y = \frac{\cos x}{1 + \sin x} \), the Quotient Rule is your go-to strategy. This nifty rule helps you find the derivative of functions that are divided by each other.

Here's how the Quotient Rule works: if you have a function \( y = \frac{u(x)}{v(x)} \), the derivative \( \frac{dy}{dx} \) is computed as:

• \[ \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \]

You first find the derivatives of the numerator \( u(x) \) and the denominator \( v(x) \). In our problem:

• \( u(x) = \cos x \) with \( u'(x) = -\sin x \)
• \( v(x) = 1 + \sin x \) with \( v'(x) = \cos x \)

Once you have these derivatives, plug them into the Quotient Rule formula to find the overall derivative. It's methodical and might seem a bit daunting at first, but with practice, it becomes second nature.

Trigonometric functions such as \( \sin x \) and \( \cos x \) play a fundamental role in calculus, particularly when differentiating and integrating. Each function has a specific rule for differentiation:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \(-\sin x \).

These derivatives arise frequently, and having them committed to memory is invaluable when solving calculus problems.

Our particular exercise involves differentiating \( \cos x \), which gives \(-\sin x \), and \( \sin x \), which contributes to \( v(x) \) in the exercise, giving \( \cos x \) after differentiation. Understanding how to differentiate these functions helps streamline applying more complex rules like the Quotient Rule.

After applying the Quotient Rule, the expressions often need to be simplified for clarity and conciseness. Simplification not only makes your final answer more readable but can also highlight insights into the function's behavior.

For our derivative \( \frac{-\left(1+\sin x\right)}{\left(1+\sin x\right)^2} \), observe that the numerator and part of the denominator share a common factor \((1 + \sin x)\). This commonality enables you to cancel out one term easily, leading to:

• \( -\frac{1}{1 + \sin x} \)

Understanding algebraic identities, like \( \sin^2 x + \cos^2 x = 1 \), is also crucial. In our example, using this identity allowed for further reduction of terms, making the derivative simpler.

These simplification techniques are essential in calculus to provide neat solutions that reflect actual changes in the function effectively.