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Chapter 3: Problem 3

Explain why the Quotient Rule is used to determine the derivative of \(\tan x\) and \(\cot x\)

Short Answer

Expert verified
Answer: The Quotient Rule is used because both \(\tan x\) and \(\cot x\) are expressed as ratios of two functions (\(\sin x\) and \(\cos x\)). The Quotient Rule helps find their derivatives by differentiating the numerator and denominator and combining them in a specific manner, resulting in the derivatives \(\sec^2 x\) for \(\tan x\) and \(-\csc^2 x\) for \(\cot x\).

Step by step solution

01

Derivative of \(\tan x\)

Recall that \(\tan x = \frac{\sin x}{\cos x}\). First, find the derivatives of the numerator and denominator: \((\sin x)' = \cos x\) and \((\cos x)' = -\sin x\). Then, apply the Quotient Rule to find the derivative of \(\tan x = \frac{\sin x}{\cos x}\): $$ (\tan x)' = \frac{(\sin x)'\cos x - \sin x(\cos x)'}{\cos^2 x} = \frac{\cos x\cos x - \sin x(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x $$
02

Derivative of \(\cot x\)

Recall that \(\cot x = \frac{\cos x}{\sin x}\). First, find the derivatives of the numerator and denominator: \((\cos x)' = -\sin x\) and \((\sin x)' = \cos x\). Then, apply the Quotient Rule to find the derivative of \(\cot x = \frac{\cos x}{\sin x}\): $$ (\cot x)' = \frac{(\cos x)'\sin x - \cos x(\sin x)'}{\sin^2 x} = \frac{-\sin x\sin x - \cos x\cos x}{\sin^2 x} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x $$ In conclusion, the Quotient Rule is used to determine the derivatives of \(\tan x\) and \(\cot x\) because both of these functions are expressed as ratios of two other functions (\(\sin x\) and \(\cos x\)).

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