Question

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

Answer: Yes, the functions \(f(x) = \tan^2x\) and \(g(x) = \sec^2x\) have the same derivative, which is \(2\tan x(\sec^2x)\). The difference between the two functions is constant and equal to -1, indicating a constant vertical shift between them.

Step-by-step solution

Find the derivative of \(f(x)=\tan^2x\)

Use the chain rule and the derivative of \(\tan x\). The derivative of \(\tan x\) is \(\sec^2x\). Thus, by applying the chain rule, we get: \( \frac{d}{dx}(\tan^2x) = 2\tan x \cdot \frac{d(\tan x)}{dx} = 2\tan x \cdot \sec^2x= 2 \tan x(\sec^2x) \)

Find the derivative of \(g(x)=\sec^2x\)

Use the chain rule and the derivative of \(\sec x\). The derivative of \(\sec x\) is \(\sec x\tan x\). Thus, by applying the chain rule, we get: \( \frac{d}{dx}(\sec^2x) = 2\sec x \cdot \frac{d(\sec x)}{dx} = 2\sec x\cdot(\sec x\tan x) = 2\tan x(\sec^2x) \)

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

Observe that the derivatives of both functions are equal: \( \frac{d}{dx}(f(x)) = 2\tan x(\sec^2x) = \frac{d}{dx}(g(x)) \) So, the derivatives of both functions are indeed the same.

Analyze the difference between \(f(x)\) and \(g(x)\)

Consider the difference between the two functions \(f-g\): \( f(x) - g(x) = \tan^2x - \sec^2x \) Now that we know their derivatives are the same, this means they differ by a constant value. Let's find the constant: \( \sec^2x = 1+\tan^2x \) (by identity), so \( \tan^2x-\sec^2x = -1 \) Thus, the difference between the two functions is a constant, and the value is equal to -1. This means that the functions are related by a constant vertical shift.

Key concepts

Chain Rule

When finding derivatives of composite functions, the chain rule is an essential tool. It allows us to differentiate compound expressions by taking the derivative of the outer function and multiplying it by the derivative of the inner function.
Let's say we have a composite function, like \(f(g(x))\). To find its derivative, we would use the chain rule:

• Take the derivative of the outer function, \(f'(g(x))\), treating \(g(x)\) as a whole, not concerning its specific form yet.
• Multiply the result by the derivative of the inner function, \(g'(x)\).

So, the derivative of \(f(g(x))\) using the chain rule is \[f'(g(x)) imes g'(x)\]. This process can simplify complex derivatives and makes handling nested functions much more manageable. In our exercise, the application of the chain rule to both \(f(x) = \tan^2(x)\) and \(g(x) = \sec^2(x)\) demonstrates its versatility.

Trigonometric Derivatives

Understanding the derivatives of trigonometric functions like \(\tan(x)\) and \(\sec(x)\) is crucial in solving calculus problems involving trigonometric expressions. Here are the basic derivatives you need for these functions:

• The derivative of \(\tan(x)\) is \(\sec^2(x)\). This follows from the quotient rule and the identity \(\tan(x) = \frac{\sin(x)}{\cos(x)}\).
• The derivative of \(\sec(x)\) involves multiple steps: first recognize that \(\sec(x) = \frac{1}{\cos(x)}\), applying the chain rule gives the result \(\sec(x)\tan(x)\).

In our problem, we apply these derivatives directly:

• To differentiate \(\tan^2(x)\), we first recognize it as \(\tan(x)\) squared and use the chain rule, multiplying by \(2 \tan(x) \sec^2(x)\).
• For \(\sec^2(x)\), we again use the chain rule, multiplying by \(2 \tan(x) \sec^2(x)\).

This shows how closely related these trig functions are and how their derivatives create similar forms.

Function Difference

When two functions have identical derivatives, this hints they are the same up to a constant. This is due to the fundamental theorem of calculus, which states that integrating identical derivatives gives the original functions differing by a constant.In our exercise, we explore the functions \(f(x) = \tan^2(x)\) and \(g(x) = \sec^2(x)\), both having the derivative \(2\tan(x)\sec^2(x)\). As a result, their difference is governed by this constant principle.

• By calculating \(f(x) - g(x)\), which equals \(\tan^2(x) - \sec^2(x)\), we reduce the expression.
• Using the trigonometric identity \(\sec^2(x) = 1 + \tan^2(x)\), the difference simplifies to a constant, which is \(-1\).

Thus, \(f(x)\) and \(g(x)\) are indeed related by this constant, reflecting a vertical shift in their graphical representation, and confirming the steadfast nature of calculus in connecting derivatives and function differences.