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Chapter 3: Problem 4

How can you use the derivatives \(\frac{d}{d x}(\sin x)=\cos x\) \(\frac{d}{d x}(\tan x)=\sec ^{2} x,\) and \(\frac{d}{d x}(\sec x)=\sec x \tan x\) to remember the derivatives of \(\cos x, \cot x,\) and \(\csc x ?\)

Short Answer

Expert verified
Question: Using the given derivatives, find the derivatives of \(\cos x\), \(\cot x\), and \(\csc x\). Given: \(\frac{d}{d x}(\sin x)=\cos x\), \(\frac{d}{d x}(\tan x)=\sec^2 x\) Answer: \(\frac{d}{d x}(\cos x) = -\sin x\), \(\frac{d}{d x}(\cot x) = -\csc^2 x\), \(\frac{d}{d x}(\csc x) = -\cot x \csc x\)

Step by step solution

01

Derivative of \(\cos x\) using the derivative of \(\sin x\)

We have the given derivative of \(\sin x\): \(\frac{d}{d x}(\sin x)=\cos x\). To obtain the derivative of \(\cos x\), we can use the trigonometric property \(\cos x = \sin(90^\circ - x)\) and apply the chain rule as follows: \(\frac{d}{d x}(\cos x) = \frac{d}{d x}(\sin (90^\circ - x))\) By applying the chain rule, we have: \(\frac{d}{d x}(\cos x) = \frac{d}{d x}[\sin (90^\circ - x)] * (-1)\) Now, using the given derivative of \(\sin x\): \(\frac{d}{d x}(\cos x) = \cos (90^\circ - x) * (-1)\) Finally, we arrive at the result: \(\frac{d}{d x}(\cos x) = \boxed{-\sin x}\)
02

Derivative of \(\cot x\) using the derivative of \(\tan x\)

We have the given derivative of \(\tan x\): \(\frac{d}{d x}(\tan x)=\sec^2 x\). To obtain the derivative of \(\cot x\), we can use the trigonometric property \(\cot x = \frac{1}{\tan x}\) and apply the quotient and chain rules as follows: \(\frac{d}{d x}(\cot x) = \frac{d}{d x}\left(\frac{1}{\tan x}\right)\) By applying the quotient and chain rules, we have: \(\frac{d}{d x}(\cot x) = \frac{-1 * \frac{d}{d x}(\tan x)}{(\tan x)^2}\) Now, using the given derivative of \(\tan x\): \(\frac{d}{d x}(\cot x) = \frac{-\sec^2 x}{(\tan x)^2}\) Finally, using the trigonometric property \(\sec^2 x = 1 + \tan^2 x\), we can simplify the result: \(\frac{d}{d x}(\cot x) = \boxed{-\csc^2 x}\)
03

Derivative of \(\csc x\) using the derivative of \(\sin x\) and \(\sec x\)

We have the given derivatives of \(\sin x\): \(\frac{d}{d x}(\sin x)=\cos x\), and \(\frac{d}{d x}(\sec x)=\sec x \tan x\). To obtain the derivative of \(\csc x\), we can use the trigonometric property \(\csc x = \frac{1}{\sin x}\) and apply the quotient and chain rules as follows: \(\frac{d}{d x}(\csc x) = \frac{d}{d x}\left(\frac{1}{\sin x}\right)\) By applying the quotient and chain rule, we have: \(\frac{d}{d x}(\csc x) = \frac{-1 * \frac{d}{d x}(\sin x)}{(\sin x)^2}\) Now, using the given derivative of \(\sin x\): \(\frac{d}{d x}(\csc x) = \frac{-\cos x}{(\sin x)^2}\) Finally, using the trigonometric identity \(\cot x \csc x = \frac{\cos x}{\sin x}\), we can simplify the result: \(\frac{d}{d x}(\csc x) = \boxed{-\cot x \csc x}\)

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Key Concepts

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

Derivative of Sine Function
The derivative of the sine function is a fundamental concept in calculus, which is used extensively in various mathematical applications. Knowing this derivative is essential as it forms the basis for understanding changes in trigonometric functions. The derivative of the sine function is expressed as:\[\frac{d}{dx}(\sin x) = \cos x\]This tells us that as the angle changes, the rate of change of the sine function is given by the cosine of that angle. The result is useful because it helps us calculate other derivatives of trigonometric functions. Understanding how to find and apply this derivative will improve your ability to analyze problems involving trigonometry and calculus. By applying this knowledge, you can determine the derivatives of other related trigonometric functions, such as cosine, by using various rules like the chain rule or trigonometric identities.
Chain Rule in Calculus
The chain rule is a powerful tool in calculus, allowing us to differentiate compositions of functions. When dealing with trigonometric functions, it's particularly handy for finding derivatives when a function is nested within another. The chain rule states:\[\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)\]This means that to differentiate a composite function, we need to take the derivative of the outer function (with the inner function as the variable) and multiply it by the derivative of the inner function.For example, when finding the derivative of \(\cos x\) using the derivative of \(\sin x\), we consider \(\cos x\) as a transformation of \(\sin x\). Applying the chain rule, we differentiate \(\sin(90^\circ - x)\) by first evaluating its derivative as a sine function, then multiplying by the derivative of the inside function, which is \(-1\). The chain rule elegantly captures these layers of function composition and helps simplify the calculus involved.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables. These identities are incredibly useful for simplifying the process of finding derivatives of trigonometric functions.Key trigonometric identities include:
  • \(\cos^2 x + \sin^2 x = 1\)
  • \(1 + \tan^2 x = \sec^2 x\)
  • \(\csc x = \frac{1}{\sin x}\)
  • \(\cot x = \frac{1}{\tan x}\)
These identities help in re-expressing trigonometric functions into forms that are easier to differentiate. For example, when computing the derivative of \(\cot x\), the identity \(1 + \tan^2 x = \sec^2 x\) helps in simplifying the expression. Similarly, identities help transform complicated expressions into simpler, more manageable forms that make the process of differentiation straightforward.Understanding and using these identities allow for a deeper insight into how various trigonometric functions interact and change with respect to each other. This knowledge is essential for success in calculus and trigonometry.

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Most popular questions from this chapter

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. A mechanical oscillator (such as a mass on a spring or a pendulum) subject to frictional forces satisfies the equation (called a differential equation) $$y^{\prime \prime}(t)+2 y^{\prime}(t)+5 y(t)=0$$ where \(y\) is the displacement of the oscillator from its equilibrium position. Verify by substitution that the function \(y(t)=e^{-t}(\sin 2 t-2 \cos 2 t)\) satisfies this equation.

The following equations implicitly define one or more functions. a. Find \(\frac{d y}{d x}\) using implicit differentiation. b. Solve the given equation for \(y\) to identify the implicitly defined functions \(y=f_{1}(x), y=f_{2}(x), \ldots.\) c. Use the functions found in part (b) to graph the given equation. \(x^{4}=2\left(x^{2}-y^{2}\right)\) (eight curve)

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.

Beginning at age \(30,\) a self-employed plumber saves \(\$ 250\) per month in a retirement account until he reaches age \(65 .\) The account offers \(6 \%\) interest, compounded monthly. The balance in the account after \(t\) years is given by \(A(t)=50,000\left(1.005^{12 t}-1\right)\) a. Compute the balance in the account after \(5,15,25,\) and 35 years. What is the average rate of change in the value of the account over the intervals \([5,15],[15,25],\) and [25,35]\(?\) b. Suppose the plumber started saving at age 25 instead of age 30\. Find the balance at age 65 (after 40 years of investing). c. Use the derivative \(d A / d t\) to explain the surprising result in part (b) and the advice: Start saving for retirement as early as possible.

86-89. Second derivatives Find \(\frac{d^{2} y}{d x^{2}}\) for the following functions. $$y=\sqrt{x^{2}+2}$$
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