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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

Expert verified
Question: Using the definition of the derivative and trigonometric identities, prove that the derivative of cos(x) is -sin(x). Answer: By applying the definition of the derivative, expanding the expression using the trigonometric identity, and simplifying the expression, we find that the derivative of cos(x) is -sin(x).

Step by step solution

01

Write down the definition of the derivative for cos(x)

We begin by writing the definition of the derivative for f(x) = cos(x): $$f'(x) = \lim_{h \to 0} \frac{\cos(x+h)-\cos(x)}{h}$$
02

Use the trigonometric identity to expand cos(x+h)

Next, we apply the trigonometric identity \(\cos(x+h) = \cos x \cdot \cos h - \sin x \cdot \sin h\) to expand the function: $$f'(x) = \lim_{h \to 0} \frac{(\cos x \cdot \cos h - \sin x \cdot \sin h)-\cos(x)}{h}$$
03

Simplify the expression

Then, we simplify the expression by distributing the subtraction: $$f'(x) = \lim_{h \to 0} \frac{\cos x \cdot \cos h - \cos x - \sin x \cdot \sin h}{h}$$
04

Factor out common terms

We can factor \(\cos x\) and \(-\sin x\) from the numerator: $$f'(x) = \lim_{h \to 0} \frac{\cos x (\cos h - 1) - \sin x (\sin h)}{h}$$
05

Use trigonometric limit identities

We split the limit as a sum of two limits and use the trigonometric limit identities: \(\lim_{h \to 0} \frac{\cos h - 1}{h}=0\) and \(\lim_{h \to 0} \frac{\sin h}{h}=1\): $$f'(x) = \cos x \cdot \lim_{h \to 0} \frac{\cos h - 1}{h} - \sin x \cdot \lim_{h \to 0} \frac{\sin h}{h}$$ $$f'(x) = \cos x \cdot 0 - \sin x \cdot 1$$
06

Write the final result

Finally, our expression simplifies to the derivative of cos(x): $$f'(x) = -\sin x$$ Hence, the derivative of \(\cos x\) is \(-\sin x\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
The derivative is a core concept in calculus, representing how a function changes as its input changes. Essentially, it measures the "slope" or incline of a function at any given point. For functions like \(f(x) = \cos(x)\), finding the derivative helps us understand how \(\cos(x)\) varies with small changes in \(x\). When working with derivatives, we often use the limit process. This involves finding the rate of change as the change in the input approaches zero, giving us the instantaneous rate of change.

In mathematical terms, the definition of the derivative of a function \(f(x)\) at a point \(x\) is given by:
  • \(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)
The notation \(f'(x)\) denotes the derivative of \(f(x)\). This expression represents the formula needed to compute the derivative, providing the foundation for solving problems like finding \(\frac{d}{dx}(\cos x)\).
Trigonometric Identities
Trigonometric identities are fundamental tools that help simplify expressions involving trigonometric functions. In the context of calculus, these identities are often used to manipulate and work with trigonometric functions effectively. One important identity used in calculus is:
  • \(\cos(x+h) = \cos x \cdot \cos h - \sin x \cdot \sin h\)
This identity decomposes the cosine of a sum of two angles into simpler components. In derivatives involving \(\cos(x)\), such identities play a crucial role in expanding and simplifying expressions to reach the derivative.

By using trigonometric identities, we can break down complex expressions and make the application of differentiation rules more straightforward. These identities also aid us in understanding the relationships between different trigonometric functions.
Limits
Limits are foundational in calculus and serve as the backbone for defining derivatives. A limit tells us what value a function approaches as the input approaches a certain point. Understanding limits is critical when dealing with derivatives because they help us determine the value that a function's instantaneous rate of change approaches as the change in input becomes infinitely small.

For the derivative of \(\cos(x)\), we look at limits such as \(\lim_{h \to 0} \frac{\cos(x+h)-\cos(x)}{h}\). By applying limit properties, particularly those for trigonometric functions, we obtain meaningful derivatives. Some trigonometric limits vital in derivative calculations include:
  • \(\lim_{h \to 0} \frac{\cos h - 1}{h}=0\)
  • \(\lim_{h \to 0} \frac{\sin h}{h}=1\)
These limit identities simplify the derivative expression and make calculus operations involving trigonometric functions feasible and precise.
Differentiation Rules
Differentiation rules, such as the power rule, product rule, quotient rule, and chain rule, are procedures applied to find the derivatives of various functions. These rules simplify the process of taking derivatives, especially in complex functions where manual application of the limit definition can be cumbersome. In the case of trigonometric functions, specific differentiation rules apply.

For the derivative of \(\cos(x)\), the differentiation process involves some known trigonometric derivative formulas:
  • The derivative of \(\cos(x)\) is \(-\sin(x)\).
Applying these rules allows us to compute derivatives quickly and efficiently, providing a structured approach to solving problems related to function changes. Differentiation rules leverage the limit definition but give us a toolkit for rapidly finding derivatives in standardized situations, like differentiating trigonometric functions.

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Most popular questions from this chapter

Find the equation of the line tangent to \(y=e^{2 x}\) at \(x=\frac{1}{2} \ln 3\)

Vibrations of a spring Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. Use equation (4) to answer the following questions. a. The period \(T\) is the time required by the mass to complete one oscillation. Show that \(T=2 \pi \sqrt{\frac{m}{k}}\) b. Assume \(k\) is constant and calculate \(\frac{d T}{d m}\) c. Give a physical explanation of why \(\frac{d T}{d m}\) is positive.

Tangency question It is easily verified that the graphs of \(y=x^{2}\) and \(y=e^{x}\) have no point of intersection (for \(x>0\) ), while the graphs of \(y=x^{3}\) and \(y=e^{x}\) have two points of intersection. It follows that for some real number \(2 < p < 3,\) the graphs of \(y=x^{p}\) and \(y=e^{x}\) have exactly one point of intersection (for \(x > 0) .\) Using analytical and/or graphical methods, determine \(p\) and the coordinates of the single point of intersection.

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure that they actually lie on the curve. Confirm your results with a graph. $$x^{2}(y-2)-e^{y}=0$$

a. Calculate \(\frac{d}{d x}\left(x^{2}+x\right)^{2}\) using the Chain Rule. Simplify your answer. b. Expand \(\left(x^{2}+x\right)^{2}\) first and then calculate the derivative. Verify that your answer agrees with part (a).

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