Yes, the derivative of $\sec x$ with respect to $x$ is indeed $\sec x \tan x$. Let's go through the reasoning behind it:

Step-by-Step Derivation:

1. Start with the function: $f(x) = \sec x = \frac{1}{\cos x}$

2. Apply the quotient rule to differentiate $\sec x$, where: $\frac{d}{dx} \left( \frac{1}{\cos x} \right) = \frac{0 \cdot \cos x - 1 \cdot (-\sin x)}{\cos^2 x}$ This simplifies to: $\frac{\sin x}{\cos^2 x}$

3. Recognize the trigonometric identity: $\frac{\sin x}{\cos^2 x} = \sec x \tan x$

Therefore: $\frac{d}{dx} \sec x = \sec x \tan x$

Would you like more details or have any questions?

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Relative Questions:

1. What is the derivative of $\tan x$?
2. How do you apply the chain rule to trigonometric functions like $\sec(2x)$?
3. Can you prove the derivative of $\csc x$ using similar steps?
4. How does the product rule apply to a function like $\sec x \cdot \cos x$?
5. What is the second derivative of $\sec x$?

Tip:

When dealing with derivatives of trigonometric functions, it's helpful to remember their identities and corresponding derivatives, as they often simplify the process!