Problem 12
Question
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 $$ [\cos x]^{\prime}=-\sin x $$
Step-by-Step Solution
Verified Answer
The derivative of \(\cos x\) is \(-\sin x\) when \(\cos x \neq 0\).
1Step 1: Differentiate Both Sides of Trigonometric Identity
Start with the well-known trigonometric identity: \( \sin^2 x + \cos^2 x = 1 \). Differentiate both sides with respect to \( x \). The derivative of the left side is \( 2\sin x \cdot \cos x + 2\cos x \cdot \frac{d}{dx}(\cos x) \), noting the use of the chain rule. The derivative of the right side is 0 because the derivative of a constant is zero.
2Step 2: Set Up the Equation from Derivatives
From Step 1, we have the equation: \( 2\sin x \cdot \cos x + 2\cos x \cdot [\cos x]' = 0 \). This equation is derived from setting the derivative of the left side of the identity equal to the derivative of the right side (which is 0).
3Step 3: Simplify the Equation
Divide every term in the equation \( 2\sin x \cdot \cos x + 2\cos x \cdot [\cos x]' = 0 \) by \( 2\cos x \), assuming \( \cos x eq 0 \). This simplification yields: \( \sin x + [\cos x]' = 0 \).
4Step 4: Solve for the Derivative of Cosine
From \( \sin x + [\cos x]' = 0 \), isolate \([\cos x]'\) to find the derivative of \( \cos x \). Subtract \( \sin x \) from both sides to obtain \([\cos x]' = -\sin x\). Hence, when \(\cos x eq 0\), the derivative of \(\cos x\) with respect to \(x\) is \(-\sin x\).
Key Concepts
Implicit DifferentiationTrigonometric IdentitiesChain RuleCalculus
Implicit Differentiation
Implicit differentiation is a technique in calculus used to find derivatives of equations not solved explicitly for one variable. Take the trigonometric identity \( \sin^2 x + \cos^2 x = 1 \). To differentiate both sides, we treat the equation as a whole. This approach lets us find derivatives even if we cannot isolate a variable on one side. For example, the derivative of the left side becomes \( 2\sin x \cdot \cos x + 2\cos x \cdot \frac{d}{dx}(\cos x) \). Here, we implicitly differentiate \( \cos x \) with respect to \( x \) because \( \cos^2 x \) depends on \( x \).
Using implicit differentiation helps solve derivatives conveniently. It is useful when working with equations where separating variables is difficult. Consequently, this method allows exploration of complex relationships without requiring explicit functions.
Using implicit differentiation helps solve derivatives conveniently. It is useful when working with equations where separating variables is difficult. Consequently, this method allows exploration of complex relationships without requiring explicit functions.
Trigonometric Identities
Trigonometric identities are equations that hold true for all values of the variables involved. A classic one is \( \sin^2 x + \cos^2 x = 1 \), which connects the sine and cosine of the same angle. This identity is pivotal in trigonometry and calculus because it simplifies expressions and assists in finding derivatives of trigonometric functions.
When computing the derivative of \( \cos x \) using the identity, it shows how these identities provide a backbone for further calculus manipulations. You can simplify expressions and find connections between trigonometric functions that are not obvious at first.
In calculus, these identities often come into play when differentiating or integrating trigonometric functions, and they streamline complex problems into more manageable parts.
When computing the derivative of \( \cos x \) using the identity, it shows how these identities provide a backbone for further calculus manipulations. You can simplify expressions and find connections between trigonometric functions that are not obvious at first.
In calculus, these identities often come into play when differentiating or integrating trigonometric functions, and they streamline complex problems into more manageable parts.
Chain Rule
The chain rule in calculus is a formula for computing the derivative of a composite function. When you have a function inside another function, the chain rule directs how to differentiate it. For instance, when you differentiate \( \cos^2 x \), you're dealing with a composite function \( f(g(x)) \) where \( f(u) = u^2 \) and \( g(x) = \cos x \).
The chain rule states: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\). Apply this when differentiating \( \cos^2 x \), which leads to: \( 2\cos x \cdot (-\sin x) \), because the derivative of \( \cos x \) is \(-\sin x\).
This rule is fundamental in calculus for differentiating complex expressions. It provides a systematic way to tackle derivatives that involve nested functions, expanding the scope of problems you can solve efficiently.
The chain rule states: \((f(g(x)))' = f'(g(x)) \cdot g'(x)\). Apply this when differentiating \( \cos^2 x \), which leads to: \( 2\cos x \cdot (-\sin x) \), because the derivative of \( \cos x \) is \(-\sin x\).
This rule is fundamental in calculus for differentiating complex expressions. It provides a systematic way to tackle derivatives that involve nested functions, expanding the scope of problems you can solve efficiently.
Calculus
Calculus is the study of how things change, focusing on rates of change and accumulation. It is divided into differential and integral calculus. Differential calculus involves derivatives, which are tools that allow us to determine how a quantity changes over time.
In the derivation exercise, we explore differential calculus by finding the derivative of \( \cos x \). This involves understanding how small changes in \( x \) affect \( \cos x \). Techniques like implicit differentiation, chain rule, and familiarity with trigonometric identities all come into play.
These tools grant the ability to dissect and analyze dynamic systems. Calculus is foundational in fields ranging from physics to engineering where understanding the behavior of changing systems is crucial.
In the derivation exercise, we explore differential calculus by finding the derivative of \( \cos x \). This involves understanding how small changes in \( x \) affect \( \cos x \). Techniques like implicit differentiation, chain rule, and familiarity with trigonometric identities all come into play.
These tools grant the ability to dissect and analyze dynamic systems. Calculus is foundational in fields ranging from physics to engineering where understanding the behavior of changing systems is crucial.
Other exercises in this chapter
Problem 8
The complete solution to Exercise 7.4 .7 is given in the Solutions Section on the web. Try your hand with showing that $$ y(t)=e^{-b t} \sin \left(\sqrt{c-b^{2}
View solution Problem 9
The derivative of \(y=\cos x\) is defined by $$ [\cos x]^{\prime}=\lim _{h \rightarrow 0} \frac{\cos (x+h)-\cos x}{h} $$ Make a plot of $$ y=-\sin t \quad \text
View solution Problem 13
How is the preference in calculus for radian measure of angles for trigonometric functions similar to the preference of the number \(e\) as the base of exponent
View solution Problem 7
At least, show that $$ y(t)=e^{-b t} \cos \left(\sqrt{c-b^{2}} t\right) $$ solves $$ y^{\prime \prime}(t)+2 b y^{\prime}(t)+c y(t)=0 \quad \text { for } \quad c
View solution