Question

Proof of \(\frac{d}{d x}(\cos x)=-\sin x\) Use the definition of the derivative and the trigonometric identity $$\cos (x+h)=\cos x \cos h-\sin x \sin h$$ to prove that \(\frac{d}{d x}(\cos x)=-\sin x.\)

Short answer

In this problem, you were asked to find the derivative of cos(x) using the definition of the derivative and the trigonometric identity. Following a step by step solution, you used the trigonometric identity to simplify the expression, applied standard limits, and arrived at the conclusion that the derivative of cos(x) is indeed -sin(x): $$\frac{d}{dx}(\cos x) = -\sin x$$.

Step-by-step solution

Write down the definition of the derivative

Recall that the derivative of a function f(x) is defined as follows: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$. We need to find the derivative of \(\cos(x)\) - which we will denote as \(f(x) = \cos(x)\).

Apply the definition of the derivative to the given function

We have \(f(x) = \cos(x)\), so we can write down the definition of the derivative for this function: $$f'(x)=\lim_{h\to 0}\frac{\cos(x+h)-\cos(x)}{h}$$.

Use the trigonometric identity

Now, we need to use the given trigonometric identity: $$\cos(x+h) = \cos x\cos h-\sin x\sin h$$. Replace \(\cos(x+h)\) in the expression for the derivative with the given identity: $$f'(x) = \lim_{h \to 0} \frac{\cos x\cos h-\sin x\sin h-\cos(x)}{h}$$.

Separate the terms in the numerator

We now separate the terms in the numerator, creating two fractions: $$f'(x)=\lim_{h\to 0}\frac{\cos x\cos h-\cos x}{h}+\lim_{h\to 0}\frac{-\sin x\sin h}{h}$$.

Factor out constants from the limits

Factor out the constants \(\cos x\) and \(-\sin x\) from the limits: $$f'(x) = \cos x \lim_{h\to 0}\frac{\cos h-1}{h} -\sin x\lim_{h\to 0}\frac{\sin h}{h}$$.

Apply standard limits

We can recognize two standard limits: $$\lim_{h\to 0}\frac{\sin h}{h} = 1$$ and $$\lim_{h\to 0}\frac{\cos h - 1}{h} = 0$$. Apply these limits to the expression: $$f'(x)=\cos x (0)-\sin x (1)=-\sin x$$.

Conclusion

We have successfully proven that, using the definition of the derivative and the given trigonometric identity, the derivative of \(\cos x\) is indeed \(-\sin x\): $$\frac{d}{dx}(\cos x) = -\sin x$$.

Key concepts

Definition of the Derivative

Understanding the derivative of a function is a foundational concept in calculus, and it represents the rate at which a function is changing at any given point. In more formal terms, the derivative of a function at a point is the limit of the average rate of change of the function over an interval as the interval length approaches zero.

The mathematical definition of a derivative is written as: \[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \].

This equation is the backbone of differentiation and is used to find the slope of the tangent line to the curve of the function at any point. In the context of the problem where we want to find the derivative of \(\cos(x)\), \(f(x) = \cos(x)\) and the derivative \(f'(x)\) represents the instantaneous rate of change of \(\cos(x)\). It's like finding the speedometer reading of a car's speed which varies over time at a specific instant, rather than calculating its average speed over a journey.

Trigonometric Identities

Trigonometric identities are equations that describe the relationships between the trigonometric functions and are used to simplify expressions or prove other mathematical propositions. One such identity, which is crucial in understanding the derivative of the cosine function, is \[ \cos(x+h) = \cos(x)\cos(h) - \sin(x)\sin(h) \].

This particular identity showcases how the cosine of a sum of two angles can be expressed in terms of the product of their individual cosines and sines. In our exercise, this identity allows us to transform the expression \(\cos(x+h)\) to a form that facilitates the application of the definition of the derivative. Understanding and wielding these identities is like having a Swiss army knife for tackling many problems in calculus, especially when dealing with trigonometric functions.

Limit of a Function

The concept of a limit in mathematics is essential for the study of calculus. It describes the idea of approaching a specific value and is the fundamental tool used for defining both the derivative and the integral. In the context of derivatives, limits help us understand the behavior of a function as it gets close to a certain point.

For example, in the problem at hand, the use of limits such as \[ \lim_{h\to 0} \frac{\sin h}{h} = 1 \] and \[ \lim_{h\to 0} \frac{\cos h - 1}{h} = 0 \] are necessary to conclude the derivative of \(\cos(x)\). These limits represent the behavior of \(\sin(h)\) and \(\cos(h) - 1\) as \(h\) approaches zero, which are essential elements in calculating the derivative correctly.

Grasping the concept of limits not only aids in performing differentiation but also in understanding the continuous nature of functions and the idea of an infinitely small change which is at the heart of calculus.