To find the derivative of the function $y = \sec(x) \csc(3x)$, we will use the product rule. The product rule states that if $y = u(x) \cdot v(x)$, then:

$\frac{dy}{dx} = u'(x) v(x) + u(x) v'(x)$

Here, let: $u(x) = \sec(x), \quad v(x) = \csc(3x)$

Step 1: Find the derivatives of $u(x)$ and $v(x)$

1. Derivative of $u(x) = \sec(x)$:

$u'(x) = \frac{d}{dx} (\sec(x)) = \sec(x) \tan(x)$

2. Derivative of $v(x) = \csc(3x)$:

$v'(x) = \frac{d}{dx} (\csc(3x))$

To differentiate $\csc(3x)$, we use the chain rule:

$v'(x) = -\csc(3x) \cot(3x) \cdot \frac{d}{dx} (3x) = -3 \csc(3x) \cot(3x)$

Step 2: Apply the product rule

Now, applying the product rule:

$\frac{dy}{dx} = u'(x) v(x) + u(x) v'(x)$

Substitute the values of $u(x), u'(x), v(x),$ and $v'(x)$:

$\frac{dy}{dx} = (\sec(x) \tan(x)) \cdot (\csc(3x)) + (\sec(x)) \cdot (-3 \csc(3x) \cot(3x))$

Simplify the expression:

$\frac{dy}{dx} = \sec(x) \tan(x) \csc(3x) - 3 \sec(x) \csc(3x) \cot(3x)$

Step 3: Factor the common terms

You can factor out $\sec(x) \csc(3x)$:

$\frac{dy}{dx} = \sec(x) \csc(3x) \left( \tan(x) - 3 \cot(3x) \right)$

Final Answer

$\frac{dy}{dx} = \sec(x) \csc(3x) \left( \tan(x) - 3 \cot(3x) \right)$

Would you like more details or have any questions?

5 Related Questions

1. How do you differentiate other trigonometric functions like $\sin(x)$ and $\cos(x)$?
2. What is the chain rule and how does it apply to trigonometric functions?
3. How do you find the derivative of the product of more than two functions?
4. How do you handle the differentiation of trigonometric functions involving composite angles, like $\sin(2x)$ or $\cos(3x)$?
5. What is the significance of trigonometric derivatives in real-world applications?

Tip

Remember that understanding the differentiation rules for trigonometric functions can help simplify many complex calculus problems!