Question

Prove in a similar manner that $$ D_{x} \cos x=-\sin x $$

Short answer

Question: Prove that the derivative of the cosine function with respect to x is equal to the negative of the sine function. Answer: To prove that the derivative of the cosine function with respect to x is equal to the negative of the sine function, we used the limit definition of the derivative and the trigonometric identities, such as the sum of angles formula for sine and cosine. After applying these techniques and evaluating the limits, we arrived at the result: $$ D_x \cos x = -\sin x $$

Step-by-step solution

Write down the limit definition of the derivative

We will start with the limit definition of the derivative for a function f(x), which is given by: $$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$ In this case, our function is the cosine function, so we will substitute f(x) with cos(x).

Substitute the function and apply the cosine sum of angles formula

Substitute f(x) with cos(x) in the limit definition and apply the sum of angles formula for cosine: $$ D_x \cos x = \lim_{h \to 0} \frac{\cos (x+h) - \cos x}{h} $$ Using the sum of angles formula for cosine, we get: $$ D_x \cos x = \lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h} $$

Simplify and separate terms

Now, we will simplify the expression and separate the terms in the numerator: $$ D_x \cos x = \lim_{h \to 0} \frac{\cos x (\cos h - 1) - \sin x \sin h}{h} $$

Divide each term by h and take the limit of each term separately

Now, we can divide each term in the numerator by h individually and take the limit separately for each term: $$ D_x \cos x = \lim_{h \to 0} [\cos x \cdot \frac{\cos h - 1}{h} - \sin x \cdot \frac{\sin h}{h}] $$

Evaluate the limits

Recall two important limits: $$ \lim_{h \to 0} \frac{\cos h - 1}{h} = 0 $$ and $$ \lim_{h \to 0} \frac{\sin h}{h} = 1 $$ Substitute these limits back into the expression: $$ D_x \cos x = \cos x \cdot 0 - \sin x \cdot 1 = -\sin x $$ Thus, we have shown that $$ D_x \cos x = -\sin x $$

Key concepts

Limit Definition of Derivative

To start understanding how we find the derivative of the cosine function, we need to recap the limit definition of the derivative. The derivative gives us the rate at which a function is changing at any given point. For any function \( f(x) \), its derivative \( f'(x) \) can be expressed through the limit definition:

• \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \)

In our case, we want to find the derivative of \( \cos x \). Hence, we replace \( f(x) \) with \( \cos x \) in the equation above. This setup is crucial because it forms the initial step in our process to prove that the derivative of \( \cos x \) is \( -\sin x \). Taking small steps through this definition, we can then build upon our foundation by incorporating other mathematical tools.
This specific setup is about calculating the changes in \( \cos x \) as \( x \) changes slightly by an amount \( h \). As \( h \) approaches zero, it helps pinpoint the exact slope of the function \( \cos x \) at \( x \). So fundamentally, the limit definition of derivative is our starting point in the quest to understand changes at the infinitesimal scale.

Trigonometric Identities

Next, let's focus on using trigonometric identities to simplify complex expressions. When dealing with trigonometric functions like cosine, these identities play a seminal role in easing calculations. One of the pivotal identities that we'll utilize is the cosine sum of angles identity:

• \( \cos(a + b) = \cos a \cos b - \sin a \sin b \)

When applying this identity to derive \( D_x \cos x \), we evaluate \( \cos(x + h) \). By substituting \( x \) with \( a \) and \( h \) with \( b \), the cosine sum of angles identity allows us to break down \( \cos(x + h) \) into smaller parts. This is invaluable as it enables us to separate the expression into components more easily handled in limits.
This identity showcases how trigonometric functions can interact and transform, giving us tools to simplify our derivative process. Not only does it simplify our function \( \cos(x+h) \), but it also exposes how these identities fit together in the bigger picture of calculus problems. It's like having a shortcut to navigate complex trigonometric landscapes, making the breakdown of variables and functions far more manageable.

Proof Techniques in Calculus

Calculus is rich with techniques to prove assertions like finding derivatives. A systematic approach can demystify even daunting tasks. Our goal here is proving that \( D_x \cos x = -\sin x \). One of the essential proof techniques applied in this context is evaluating limits of separated terms.Let’s focus on the expression after substituting and simplifying using trigonometric identities:

• \( D_x \cos x = \lim_{h \to 0} [\cos x \cdot \frac{\cos h - 1}{h} - \sin x \cdot \frac{\sin h}{h}] \)

Here, we treat each term in the limit separately. This approach is powerful.

• First, \( \lim_{h \to 0} \frac{\cos h - 1}{h} = 0 \). With intuition, this might be less obvious but can be confirmed through advanced limit evaluation simply by graphing \( \cos h \) around 0.
• Next, the limit \( \lim_{h \to 0} \frac{\sin h}{h} = 1 \) is a foundational result, often taught alongside the basics of derivatives tied to the sine function.

By plugging these evaluations back, we conclude that the derivative of cosine is indeed \( -\sin x \). This separation of terms and subsequent evaluation allows us to harness pre-proven limits, like building blocks to construct our final result. Such techniques are instrumental in calculus, providing both a structured method and a deeper understanding of mathematical behavior.