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Chapter 4: Problem 37

Verify that the functions \(f(x)=\tan ^{2} x\) and \(g(x)=\sec ^{2} x\) have the same derivative. What can you say about the difference \(f-g ?\) Explain.

Short Answer

Expert verified
In conclusion, the derivatives of functions f(x) and g(x) have the same magnitude but opposite signs. The difference between the functions f(x) and g(x) is always -1, which is a constant. This indicates that while the rate of change of these functions has the same magnitude, they change in opposite directions, and the functions themselves are off by a constant factor.

Step by step solution

01

Find the derivative of f(x)

To find \(f'(x)\), we use the chain rule for \(\tan^2x\). Recall, the chain rule is given by: \(\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\). So, \(f'(x) = \frac{d(\tan^2x)}{dx} = 2\tan(x)\cdot\frac{d(\tan x)}{dx}\). Now we need to find the derivative of \(\tan x\). We know that \(\tan x = \frac{\sin x}{\cos x}\). Using the quotient rule, we get: \(\frac{d(\tan x)}{dx} = \frac{\cos x \cdot \frac{d(\sin x)}{dx} - \sin x \cdot \frac{d(\cos x)}{dx}}{\cos^2x} = \frac{\cos^2x+\sin^2x}{\cos^2x} = \sec^2x\) So, \(f'(x) = 2\tan(x) \cdot \sec^2(x)\).
02

Find the derivative of g(x)

To find \(g'(x)\), we use the chain rule for \(\sec^2x\). \(g'(x) = \frac{d(\sec^2x)}{dx} = 2\sec(x)\cdot\frac{d(\sec x)}{dx}\). Now we need to find the derivative of \(\sec x\). We know that \(\sec x = \frac{1}{\cos x}\). Using the chain rule, we get: \(\frac{d(\sec x)}{dx} = -\sec x\cdot\tan x\) So, \(g'(x) = 2\sec(x) \cdot -\sec x \cdot \tan x = -2\sec^2(x) \cdot \tan(x)\).
03

Compare the derivatives of f(x) and g(x)

We have found the derivatives of f(x) and g(x): \(f'(x) = 2\tan(x)\sec^2(x)\) \(g'(x) = -2\sec^2(x)\tan(x)\) Both derivatives are the same, except for the negative sign.
04

Analyze the difference f(x)-g(x) and conclude

Now let's analyze the difference between the functions f(x) and g(x): \(f(x)-g(x) = \tan^2(x) - \sec^2(x) = \frac{\sin^2(x)}{\cos^2(x)} - \frac{1}{\cos^2(x)} = \frac{\sin^2(x) - 1}{\cos^2(x)} = \frac{-(1-\sin^2(x))}{\cos^2(x)} = \frac{-\cos^2(x)}{\cos^2(x)} = -1\) The difference between f(x) and g(x) is always -1, which is a constant. This means that the functions f(x) and g(x) are off by a constant factor, and their derivatives have the same magnitude but opposite sign.

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