91Ó°ÊÓ

The chain rule is a fundamental technique in calculus used to differentiate composite functions. To apply this rule, you first identify the inner function and the outer function. Here's an easy way to remember it: if you have a function inside another function, you will use the chain rule.
In our problem, the function is \( y = \cot(3x - 1) \). Here, the outer function is \( \cot(u) \) and the inner function is \( u = 3x - 1 \).
The chain rule tells us that to differentiate a composite function, you first differentiate the outer function, keeping the inner function the same, and then multiply this by the derivative of the inner function.
So, in our example:

• The derivative of \( \cot(u) \) is \ -\csc^2(u) \.
• Multiply this by the derivative of \( u = 3x - 1 \), which is \( 3 \).

Hence, \ \frac{dy}{dx} = -3 \csc^2(3x - 1) \.

The product rule is used to differentiate functions that are products of two or more functions. The key formula you need to remember is:
\ \frac{d}{dx}[uv] = u \frac{dv}{dx} + v \frac{du}{dx} \ where \( u \) and \( v \) are functions of \( x \).
In our problem, to find the second and third derivatives, we used the product rule combined with the chain rule:
For instance, for \ \frac{d^2y}{dx^2} \ where \ u = \csc^2(u) \ and \ v = \cot(u) \, we have:

• \( \ u = -3 \csc^2(3x - 1) \)
• \( \ v = \cot(3x - 1) \)

Differentiating these, we use the product rule to handle each term separately, ensuring that we apply the correct differentiation technique to each part. In general, for more complicated differentiation, combining both the chain rule and product rule sets you up for success.

Differentiation involves several methods to find the derivatives of functions. Here are the primary techniques that are very useful:

• **Power Rule:** Used when you have a term like \( x^n \). Its derivative is \ n x^{n-1} \.
• **Chain Rule:** This is used for composite functions as we saw earlier.
• **Product Rule:** This is used for products of functions as mentioned before.
• **Quotient Rule:** Used when you have a division of two functions: \ \frac{d}{dx}[\frac{u}{v}] = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \

For the given problem \ y = \cot(3x - 1) \, all these rules come in handy as you differentiate step-by-step. We started with the chain rule, simplified, then used the product rule for the second and third derivatives. It’s like building blocks – each technique helps you get closer to the final solution.