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Chapter 7: Problem 5

a. Use \(\cot x=\frac{\cos x}{\sin x}\) and the quotient rule to show that $$ [\cot x]^{\prime}=-\csc ^{2} x $$ b. Use \(\sec x=\frac{1}{\cos x}=(\cos x)^{-1}\) and the power chain rule to show that $$ [\sec x]^{\prime}=\sec x \tan x $$ c. Show that $$ [\csc x]^{\prime}=-\csc x \cot x . $$

Short Answer

Expert verified
Use trigonometric identities and differentiation rules to derive each expression.

Step by step solution

01

Express \( \cot x \) Using Quotient Rule

First, recall that \( \cot x = \frac{\cos x}{\sin x} \). We need to find the derivative of this ratio using the Quotient Rule: \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \). Here, \( u = \cos x \) and \( v = \sin x \).
02

Differentiate Numerator and Denominator

Differentiate \( u = \cos x \) to find \( u' = -\sin x \) and \( v = \sin x \) to find \( v' = \cos x \).
03

Apply Quotient Rule to \( \cot x \)

Plug the derivatives into the Quotient Rule formula: \[ (-\sin x)(\sin x) - (\cos x)(\cos x) \] over \( (\sin x)^2 \), simplifying to \( \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} \).
04

Simplify using Trigonometric Identity

Recall \( \sin^2 x + \cos^2 x = 1 \). Thus, \(-\sin^2 x - \cos^2 x = -1\). Therefore, the expression simplifies to \( \frac{-1}{\sin^2 x} = -\csc^2 x \).
05

Use Power Chain Rule for \( \sec x \)

Express \( \sec x = (\cos x)^{-1} \). Use the Power Chain Rule: \( \frac{d}{dx}[f(x)]^n = -n[f(x)]^{n-1}f'(x) \) with \( n = -1 \) and \( f(x) = \cos x \).
06

Differentiate \( \cos x \) and Apply Power Rule

Differentiate \( \cos x \) to get \( -\sin x \). Then, apply the Power Chain Rule to find \( (-1)(\cos x)^{-2}(-\sin x) = \sec^2 x \cdot \sin x = \sec x \tan x \).
07

Express \( \csc x \) Using Power Rule

Write \( \csc x = (\sin x)^{-1} \). Use the Power Rule: \( \frac{d}{dx} (\sin x)^{-1} = -1(\sin x)^{-2}(\cos x) \).
08

Simplify and Find Derivative of \( \csc x \)

Combine the derived components: \( \csc x \) and \( \cot x \) as \( \csc x \cdot \cot x = \frac{1}{\sin x} \cdot \frac{\cos x}{\sin x} = \frac{\cos x}{\sin^2 x} \), which simplifies to \( -\csc x \cot x \).

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Key Concepts

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

Quotient Rule
In calculus, the Quotient Rule is a method for finding the derivative of a function that is the quotient of two other functions. It's particularly useful when dealing with functions expressed as fractions. The rule states:\[\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2}.\]Here, the function \( u \) is the numerator and \( v \) is the denominator.
  • Differentiate \( u \) to find \( u' \).
  • Differentiate \( v \) to find \( v' \).
  • Substitute these into the formula.
Let's consider an example using \( \cot x = \frac{\cos x}{\sin x} \):
  • Let \( u = \cos x \) and \( v = \sin x \).
  • Then, \( u' = -\sin x \) and \( v' = \cos x \).
  • Applying the Quotient Rule leads to \( \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x \).
This confirms the derivative of \( \cot x \) is \( -\csc^2 x \). The Quotient Rule combined with trigonometric identities provides a powerful tool for differentiation.
Power Chain Rule
The Power Chain Rule is an extension of the Power Rule for derivatives, which deals with functions raised to a power, especially when these functions are themselves composite. The Power Chain Rule can be expressed as:\[\frac{d}{dx}[f(x)]^n = n[f(x)]^{n-1}f'(x).\]This implies we multiply the derivative of the base function by the exponent, reducing the exponent by one.
  • First, differentiate the outer function.
  • Next, differentiate the inner function.
  • Finally, multiply these derivatives.
For example, find the derivative of \( \sec x = (\cos x)^{-1} \):
  • Here, \( f(x) = \cos x \) and \( n = -1 \).
  • The derivative \( f'(x) = -\sin x \).
  • Applying the Power Chain Rule yields \( -1(\cos x)^{-2}(-\sin x) = \sec^2 x \cdot \sin x = \sec x \tan x \).
Thus, the derivative of \( \sec x \) is \( \sec x \tan x \). The Power Chain Rule simplifies complex derivative problems by breaking them down into manageable steps.
Trigonometric Identities
Trigonometric identities are equations involving trigonometric functions that are true for every value of the occurring variables. They play a crucial role in simplifying complex expressions and finding derivatives.For example, consider the identity \( \sin^2 x + \cos^2 x = 1 \). This identity allows us to express complex derivatives in simpler forms. Let's demonstrate using the derivative of \( \cot x \) as discussed earlier:
  • The expression \( -\sin^2 x - \cos^2 x = -1 \) utilizes this identity.
Now, let's apply these concepts to find the derivative of \( \csc x \):
  • Express \( \csc x \) as \( (\sin x)^{-1} \).
  • The Power Rule provides \(-1(\sin x)^{-2} \cdot (\cos x) = \frac{-1}{(\sin x)^2} \cdot \cos x \).
  • This simplifies to \( -\csc x \cot x \), using \( \cot x = \frac{\cos x}{\sin x} \).
Understanding these identities aids in calculus by making derivative calculations more intuitive and streamlined.

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Most popular questions from this chapter

Alternate derivation of \([\cos x]^{\prime}=-\sin x\) using implicit differentiation and \([\sin x]^{\prime}=\cos x\) Compute the derivatives of both sides of the identity $$ \sin ^{2} x+\cos ^{2} x=1 \quad \text { and obtain } \quad 2 \sin x \cos x+2 \cos x[\cos x]^{\prime}=0 $$ Use the last equation to argue that if \(\cos x \neq 0\) then $$ [\cos x]^{\prime}=-\sin x $$

Compute \(y^{\prime}\) and solve for \(t\) in \(y^{\prime}(t)=0 .\) Sketch the graphs and find the highest and the lowest points of the graphs of: a. \(y=\sin t+\cos t \quad 0 \leq t \leq \pi\) b. \(y=e^{-t} \sin t \quad 0 \leq t \leq \pi\) c. \(y=\sqrt{3} \sin t+\cos t \quad 0 \leq t \leq \pi\) d. \(y=\sin t \cos t \quad 0 \leq t \leq \pi\) e. \(y=e^{-t} \cos t \quad 0 \leq t \leq \pi\) f. \(y=e^{-\sqrt{3} t} \cos t \quad 0 \leq t \leq \pi\) Note: \(\cos ^{2} t-\sin ^{2} t=\cos (2 t)\)

Show that if \(B\) and \(\omega\) are constants and \(y(t)=B \cos (\omega t),\) then $$ y(0)=B \quad y^{\prime}(0)=0 \quad \text { and } \quad y^{\prime \prime}(t)+\omega^{2} y(t)=0 . $$

The difference quotient $$ \frac{F(t+h)-F(t)}{h} \quad \text { approximates } \quad F^{\prime}(t) $$ when \(h\) is 'small.' Make a plot of $$ y=\cos t \quad \text { and of } \quad \frac{\sin (t+0.2)-\sin t}{0.2} \quad-\frac{\pi}{2} \leq t \leq 2 \pi . $$ Repeat, using \(h=0.05\) instead of \(h=0.2\).

We need the identities $$ \begin{array}{l} \sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2} \\ \cos x-\cos y=-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2} \end{array} $$ in the next two exercises and in the next section. It is unlikely that you remember them from a trigonometry class. We hope you do remember, however, the double angle formulas 7.6 , $$ \sin (A+B)=\sin A \cos B+\cos A \sin B \quad \cos (A+B)=\cos A \cos B-\sin A \sin B $$ a. Use \(\sin (A+B)=\sin A \cos B+\cos A \sin B\) and the identities, \(\sin (-A)=-\sin A\) and \(\cos (-A)=\cos A,\) to show that $$ \sin (A-B)=\sin A \cos B-\cos A \sin B $$ b. Use the equations $$ \begin{array}{l} \sin (A+B)=\sin A \cos B+\cos A \sin B \\ \sin (A-B)=\sin A \cos B-\cos A \sin B \end{array} $$ to show that $$ \sin (A+B)-\sin (A-B)=2 \cos A \sin B $$ c. Solve for \(A\) and \(B\) in $$ \begin{array}{l} A+B=x \\ A-B=y \end{array} $$ d. Substitute the values for \(A+B, A-B, A,\) and \(B\) into Equation 7.7 to obtain $$ \sin x-\sin y=2 \cos \frac{x+y}{2} \sin \frac{x-y}{2} $$ e. Use \(\cos (A+B)=\cos A \cos B-\sin A \sin B\) to show that $$ \cos x-\cos y=-2 \sin \left(\frac{x+y}{2}\right) \sin \left(\frac{x-y}{2}\right) $$ The argument will be similar to the previous steps.
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