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To derive the formulas for the derivatives of the given functions \( \sec x \), \( \csc x \), and \( \cot x \), we will use the basic differentiation rules and trigonometric identities.

### (a) Derivative of \( \sec x \)
Recall that \( \sec x = \frac{1}{\cos x} \).

Using the quotient rule:
\[
\frac{d}{dx} \sec x = \frac{d}{dx} \left( \frac{1}{\cos x} \right)
\]
Let \( u = 1 \) and \( v = \cos x \), then:
\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}
\]
Here, \( u' = 0 \) and \( v' = -\sin x \). So:
\[
\frac{d}{dx} \sec x = \frac{0 \cdot \cos x - 1 \cdot (-\sin x)}{(\cos x)^2} = \frac{\sin x}{\cos^2 x} = \sec x \tan x
\]

### (b) Derivative of \( \csc x \)
Recall that \( \csc x = \frac{1}{\sin x} \).

Using the quotient rule:
\[
\frac{d}{dx} \csc x = \frac{d}{dx} \left( \frac{1}{\sin x} \right)
\]
Let \( u = 1 \) and \( v = \sin x \), then:
\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}
\]
Here, \( u' = 0 \) and \( v' = \cos x \). So:
\[
\frac{d}{dx} \csc x = \frac{0 \cdot \sin x - 1 \cdot \cos x}{(\sin x)^2} = \frac{-\cos x}{\sin^2 x} = -\csc x \cot x
\]

### (c) Derivative of \( \cot x \)
Recall that \( \cot x = \frac{\cos x}{\sin x} \).

Using the quotient rule:
\[
\frac{d}{dx} \cot x = \frac{d}{dx} \left( \frac{\cos x}{\sin x} \right)
\]
Let \( u = \cos x \) and \( v = \sin x \), then:
\[
\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2}
\]
Here, \( u' = -\sin x \) and \( v' = \cos x \). So:
\[
\frac{d}{dx} \cot x = \frac{(-\sin x) \sin x - \cos x \cos x}{(\sin x)^2} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x
\]

In summary:
1. \(\frac{d}{dx} \sec x = \sec x \tan x\)
2. \(\frac{d}{dx} \csc x = -\csc x \cot x\)
3. \(\frac{d}{dx} \cot x = -\csc^2 x\)

Would you like any details or have any questions?

Here are five related questions to expand on this topic:
1. What is the derivative of \(\tan x\)?
2. How can you derive the second derivative of \(\sec x\)?
3. What is the integral of \(\sec x \tan x\)?
4. Can you find the derivative of \(\sec^2 x\)?
5. How would you apply the chain rule to find the derivative of \(\sec(ax + b)\)?

**Tip:** Always remember to simplify trigonometric expressions using identities when working with derivatives and integrals to make calculations easier.

Derive the formula for the derivative with respect to x of sec x, csc x, and cot x.

Learn how to derive the formulas for the derivatives of sec x, csc x, and cot x. This solution involves using differentiation rules and trigonometric identities. Suitable for high school students in Grades 11-12.

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