Question

Verify the following derivative formulas using the Quotient Rule. $$\frac{d}{d x}(\sec x)=\sec x \tan x$$

Short answer

Question: Verify the derivative of sec x using the Quotient Rule and show that the calculated derivative matches the given formula: $\frac{d}{d x}(\sec x) = \sec x \tan x$.

Step-by-step solution

Rewrite sec x in terms of cos x

Recall that \(\sec x\) is the reciprocal of \(\cos x\), which means \(\sec x = \frac{1}{\cos x}\).

Apply the Quotient Rule

Using the Quotient Rule, we have: $$\frac{d}{d x}(\frac{u}{v}) = \frac{u'v - uv'}{v^2}$$ where \(u = 1\) and \(v = \cos x\). Now calculate the derivatives of \(u\) and \(v\): $$u' = \frac{d}{dx}(1) = 0 \hspace{10mm} \mbox{and} \hspace{10mm} v' = \frac{d}{dx}(\cos x) = -\sin x$$

Plug the derivatives into the Quotient Rule

Using the derivatives calculated above, plug the values into the Quotient Rule formula: $$\frac{d}{d x}(\sec x) = \frac{0 \cdot \cos x - 1 \cdot (-\sin x)}{(\cos x)^2}$$

Simplify the expression

Simplify the expression above as: $$\frac{\sin x}{\cos^2 x}$$ Now, recall the trigonometric identities for \(\sec x\) and \(\tan x\): $$\sec x = \frac{1}{\cos x} \hspace{10mm} \mbox{and} \hspace{10mm} \tan x = \frac{\sin x}{\cos x}$$ Using these identities, rewrite the expression as: $$\frac{\sin x}{\cos^2 x} = (\frac{1}{\cos x}) (\frac{\sin x}{\cos x}) = \sec x \tan x$$

Conclusion

We have verified that: $$\frac{d}{d x}(\sec x) = \sec x \tan x$$

Key concepts

Understanding Derivatives

In calculus, a derivative represents the rate at which a function is changing at any given point. For a function of a single variable, the derivative at a particular value gives us the slope of the tangent line to the function's graph at that point. This concept is essential for understanding the behavior of mathematical functions.

When dealing with derivatives, one of the key rules we utilize is the Quotient Rule. The Quotient Rule is particularly useful for finding the derivative of a function that is expressed as a quotient of two differentiable functions. Given two functions, \(u(x)\) and \(v(x)\), their derivative when divided can be found using:

• \(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}\)

In our example, the \'\(\sec x\)' was represented in terms of \(\cos x\), leading to an application of the Quotient Rule as \(\frac{1}{\cos x}\). This allows us to find the derivative using the relation of changes between the numerator and the denominator.

Trigonometric Identities Explained

Trigonometric identities are fundamental relationships between trigonometric functions that hold true for all angles. These identities are crucial when working with derivatives and proofs in calculus.

To approach the problem, it's helpful to understand that \(\sec x\) is defined as \(\frac{1}{\cos x}\) and \(\tan x\) is \(\frac{\sin x}{\cos x}\). These relationships not only help in simplifying expressions but also in connecting different trigonometric functions.

Having rewritten \(\sec x\) as \(\frac{1}{\cos x}\), applying the Quotient Rule directly provides us with a relationship that involves \(\sin x\) and \(\cos x\). Using identities, the expression \(\frac{\sin x}{\cos^2 x}\) can be further simplified to \(\sec x \tan x\) by re-substituting \(\sec x\) and \(\tan x\) with their definitions. This step ensures clarity and highlights the interconnectedness of trigonometric functions.

Delving into Calculus Proofs

Proofs are logical arguments that verify the truth of a mathematical statement. In calculus, proofs are used to confirm the validity of derivative formulas and other mathematical properties.

To verify the derivative of \(\sec x = \sec x \tan x\), the proof begins with rewriting \(\sec x\) using trigonometric identities and then applying calculus techniques like the Quotient Rule. Each step in the proof builds upon previous reasoning, ensuring that the transition from one expression to another is justified.

By understanding why and how each part of the problem aligns with calculus rules, students can cement their comprehension of derivative proofs. These proofs not only reinforce computational skills but also deepen understanding of the mathematical concepts underlying calculus. This skill is vital for tackling more complex calculus problems and real-world applications.