Question

Proof Prove that \(\frac{d}{d x}[\cos x]=-\sin x\)

Short answer

So, the derivative of \(\cos x\) is indeed \(-\sin x\).

Step-by-step solution

Define the derivative

Since the question is not making specific reference to a way of proving it, it's probably best to refer to the definition of a derivative from first principles: \(f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h}\).

Apply the definition to the function \(\cos x\)

So for finding the derivative of \(\cos x\), the formula becomes: \(\lim_{{h \to 0}} \frac{{\cos(x+h) - \cos(x)}}{h}\).

Use the cosine addition formula

The formula for cosine of the sum of two angles is \(\cos(a + b) = \cos a \cos b - \sin a \sin b\). Using that, the expression \(\cos(x+h)\) can be written as \(\cos x \cos h - \sin x \sin h\). So, the formula from the previous step becomes: \(\lim_{{h \to 0}} \frac{{(\cos x \cos h - \sin x \sin h) - \cos x}}{h}\).

Simplify the expression

Rewrite the fraction in the last step as two separate fractions: \(\lim_{{h \to 0}} \cos x \frac{{\cos h - 1}}{h} - \lim_{{h \to 0}} \sin x \frac{{\sin h}}{h}\).

Identify known limits

The limit of \(\frac{{\cos h - 1}}{h}\) as h approaches 0 is 0 and the limit of \(\frac{{\sin h}}{h}\) as h approaches 0 is 1. So, the expression simplifies to: \(0 \cdot \cos x - 1 \cdot \sin x\).

Finalize the result

This in turn simplifies to \(-\sin x\). Which means \(\frac{d}{d x}[\cos x]=-\sin x\)

Key concepts

First Principles of Derivatives

In calculus, the concept of derivatives is fundamental. At its core, the derivative represents how a function changes as its input changes. Imagine a car's speedometer: if the car's speed changes smoothly, the speedometer's needle is moving constantly. Similarly, the derivative tells us about changes in the function's value.

The derivative from first principles, also known as the limit definition of a derivative, provides a formal foundation. For a function \(f(x)\), the derivative \(f'(x)\) is given by the limit:

\[f'(x) = \lim_{{h \to 0}} \frac{{f(x+h) - f(x)}}{h}\]
This formula essentially computes the slope of the tangent line to the function \(f(x)\) at any point \(x\). By using smaller values of \(h\), we approach the exact slope at that point—just like the exact speed of the car at a specific moment.

The beauty of first principles lies in its ability to prove derivative formulas from the ground up, void of shortcuts or assumptions. For the cosine function, this definition transforms into a powerful tool to demonstrate its rate of change as \(-\sin x\).

Trigonometric Identities

Trigonometric identities are crucial for working with derivatives, especially when handling trigonometric functions like \(\cos x\). These identities simplify expressions and make it possible to manipulate and solve complex equations.

The key identity used in our proof is the cosine addition formula:

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

This identity allows us to express \(\cos(x+h)\) in terms of \(\cos x\), \(\cos h\), \(\sin x\), and \(\sin h\). When applying the definition of the derivative to \(\cos x\), we replace \(\cos(x+h)\) using this formula. This step reveals the underlying pattern that simplifies our expression and leads us to a solution.

Understanding and applying these trigonometric identities not only clarifies each parameter's behavior in the equation but also fosters a deeper appreciation of the interconnected nature of mathematics. It's like having a map in an unfamiliar city, helping easily spot shortcuts and essential routes.

Calculus Proof

Proving that the derivative of \(\cos x\) is \(-\sin x\) is a satisfying exploration into the world of calculus. Let's take a closer look at how each step contributes to the proof, weaving together limits and trigonometry.

Initially, we apply the first principles of derivatives to the function \(\cos x\):
\[\lim_{{h \to 0}} \frac{{\cos(x+h) - \cos(x)}}{h}\]
Then, armed with the trigonometric identity \(\cos(x+h) = \cos x \cos h - \sin x \sin h\), the expression becomes manageable:

\[\lim_{{h \to 0}} \frac{{(\cos x \cos h - \sin x \sin h) - \cos x}}{h}\]
By separating the expression into distinct parts and calculating known limits:

• \(\lim_{{h \to 0}} \cos x \frac{{\cos h - 1}}{h}\) which equals zero
• \(\lim_{{h \to 0}} \sin x \frac{{\sin h}}{h}\) which equals \(\sin x\)

We find that the derivative indeed simplifies nicely:
\[0 \cdot \cos x - 1 \cdot \sin x = -\sin x\]
This proof elegantly blends the meticulous process of calculus with the simplicity of trigonometry, illustrating how foundational principles and identities affirm the mathematical truths we accept daily.