To find the derivative [tex]\(\frac{dy}{dx}\)[/tex] of the function [tex]\(y = 3 \cos x \cot x\)[/tex], we will use the product rule and some trigonometric identities.
Given:
[tex]\[ y = 3 \cos x \cot x \][/tex]
Let's denote:
[tex]\[ u = 3 \cos x \][/tex]
[tex]\[ v = \cot x \][/tex]
Then [tex]\( y \)[/tex] can be rewritten as:
[tex]\[ y = u \cdot v \][/tex]
The product rule states that:
[tex]\[ \frac{d (u \cdot v)}{dx} = \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx} \][/tex]
First, we find [tex]\(\frac{du}{dx}\)[/tex]:
[tex]\[ u = 3 \cos x \][/tex]
[tex]\[ \frac{du}{dx} = 3 \cdot \frac{d (\cos x)}{dx} = 3 (-\sin x) = -3 \sin x \][/tex]
Next, we find [tex]\(\frac{dv}{dx}\)[/tex]:
[tex]\[ v = \cot x = \frac{\cos x}{\sin x} \][/tex]
Using the quotient rule:
[tex]\[ \frac{dv}{dx} = \frac{\sin x \cdot (-\sin x) - \cos x \cdot \cos x}{(\sin x)^2} \][/tex]
[tex]\[ \frac{dv}{dx} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} \][/tex]
Using the Pythagorean identity [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex]:
[tex]\[ \frac{dv}{dx} = \frac{-1}{\sin^2 x} = -\csc^2 x \][/tex]
Now using the product rule:
[tex]\[ \frac{dy}{dx} = \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx} \][/tex]
Substituting the values of [tex]\(\frac{du}{dx}\)[/tex], [tex]\(u\)[/tex], [tex]\(\frac{dv}{dx}\)[/tex], and [tex]\(v\)[/tex]:
[tex]\[ \frac{dy}{dx} = (-3 \sin x) \cdot \cot x + (3 \cos x) \cdot (-\csc^2 x) \][/tex]
We know that [tex]\(\cot x = \frac{\cos x}{\sin x}\)[/tex] and [tex]\(\csc x = \frac{1}{\sin x}\)[/tex], so [tex]\(\csc^2 x = \frac{1}{\sin^2 x}\)[/tex].
Therefore:
[tex]\[ \frac{dy}{dx} = -3 \sin x \cdot \frac{\cos x}{\sin x} + 3 \cos x \cdot \left( -\frac{1}{\sin^2 x} \right) \][/tex]
Simplifying:
[tex]\[ \frac{dy}{dx} = -3 \cos x - 3 \cos x \cdot (\csc^2 x) \][/tex]
Putting it all together:
[tex]\[ \frac{dy}{dx} = -3 \cos x - 3 \cos x \cdot \left( \frac{1}{\sin^2 x} \right) \][/tex]
This can be further simplified as:
[tex]\[ \frac{dy}{dx} = 3 (-\cot^2 x - 1) \cos x - 3 \sin x \cot x \][/tex]
Therefore, the final form of the derivative is:
[tex]\[ 3 (-\cot^2 x - 1) \cos x - 3 \sin x \cot x \][/tex]
