To derive the derivative of $\sec(x)$ using the quotient rule, we start by expressing $\sec(x)$ in terms of sine and cosine. Recall that:

$\sec(x) = \frac{1}{\cos(x)}$

Now, we will apply the quotient rule, which states that if $f(x) = \frac{g(x)}{h(x)}$, then:

$\frac{d}{dx} \left( \frac{g(x)}{h(x)} \right) = \frac{h(x)g'(x) - g(x)h'(x)}{(h(x))^2}$

Step 1: Set up the quotient rule

For $\sec(x) = \frac{1}{\cos(x)}$, we have:

• $g(x) = 1$
• $h(x) = \cos(x)$

Now, let's differentiate:

• $g'(x) = 0$ because the derivative of a constant (1) is 0.
• $h'(x) = \frac{d}{dx}[\cos(x)] = -\sin(x)$.

Step 2: Apply the quotient rule

Substitute these into the quotient rule formula:

$\frac{d}{dx} \left( \frac{1}{\cos(x)} \right) = \frac{\cos(x)(0) - 1(-\sin(x))}{(\cos(x))^2}$

This simplifies to:

$\frac{d}{dx} \left( \sec(x) \right) = \frac{\sin(x)}{\cos^2(x)}$

Step 3: Express in terms of secant and tangent

Next, recall that:

$\tan(x) = \frac{\sin(x)}{\cos(x)} \quad \text{and} \quad \sec(x) = \frac{1}{\cos(x)}$

Thus, we can rewrite $\frac{\sin(x)}{\cos^2(x)}$ as:

$\frac{\sin(x)}{\cos^2(x)} = \frac{1}{\cos(x)} \cdot \frac{\sin(x)}{\cos(x)} = \sec(x) \tan(x)$

Final result:

$\frac{d}{dx} \left( \sec(x) \right) = \sec(x) \tan(x)$

Thus, the derivative of $\sec(x)$ is $\sec(x) \tan(x)$, as required.

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Would you like more details on any of the steps? Here are five follow-up questions:

1. How does the quotient rule differ from the product rule in differentiation?
2. Can you explain how the derivative of $\tan(x)$ is found using the quotient rule?
3. What is the geometrical interpretation of the derivative of $\sec(x)$?
4. How can we derive the derivative of $\csc(x)$ using a similar approach?
5. What are some real-world applications of the secant function in trigonometry?

Tip: Memorizing the derivatives of basic trigonometric functions will help you solve complex problems faster!