### Video Transcript

Given that π¦ is equal to seven times the cot of three π₯ divided by three π₯ minus seven, find the first derivative of π¦ with respect to π₯.

The question wants us to find the first derivative of π¦ with respect to π₯. And we can see that π¦ is the quotient of two functions. And thereβs a few different ways of finding this derivative. For example, we could rewrite the cot of three π₯ as one divided by the tan of three π₯. However, this would make the denominator of our quotient a product of two functions. This would make finding the derivative more complicated. So, in this case, itβs easier to leave our quotient as it is and then just apply the quotient rule.

And we recall the quotient rule tells us if we have that π¦ is equal to the quotient of two functions π’ and π£, then dπ¦ by dπ₯ is equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all divided by π£ squared. So, in our case, weβll set π’ of π₯ to be the function in our numerator. Thatβs seven times the cot of three π₯. And weβll set π£ of π₯ to be the function in our denominator. Thatβs three π₯ minus seven.

To apply the quotient rule, we need to find expressions for dπ’ by dπ₯ and dπ£ by dπ₯. Letβs start by finding dπ’ by dπ₯. Thatβs the derivative of seven times the cot of three π₯ with respect to π₯. And to find the derivative of this expression, we need to recall one of our standard trigonometric derivative results. For constants π and π, the derivative of π times the cot of ππ₯ with respect to π₯ is equal to negative ππ times the csc squared of ππ₯. This is a standard result which we should commit to memory. It will make finding derivatives involving the cot of π₯ easier.

In fact, since we know the cot of ππ₯ is equivalent to the cos of ππ₯ divided by the sin of ππ₯, we can actually derive this result by applying the quotient rule to the cos of ππ₯ divided by the sin of ππ₯. Letβs now apply this result to find the derivative of our function π’ of π₯. Applying this result gives us dπ’ by dπ₯ is equal to negative three times seven csc squared of three π₯. And we can simplify this since negative three times seven is equal to negative 21. We now want to find an expression for dπ£ by dπ₯. Thatβs the derivative of three π₯ minus seven with respect to π₯. This is just the derivative of a linear function. So itβs equal to the coefficient of π₯, which in this case is three.

Weβre now ready to apply the quotient rule to find dπ¦ by dπ₯. Itβs equal to π£ times dπ’ by dπ₯ minus π’ times dπ£ by dπ₯ all divided by π£ squared. Substituting in our expressions for π’, π£, dπ’ by dπ₯, and dπ£ by dπ₯. We get that the derivative of π¦ with respect to π₯ is equal to three π₯ minus seven multiplied by negative 21 csc squared of three π₯ minus seven times the csc of three π₯ times three all divided by three π₯ minus seven squared. We can simplify this expression.

Seven multiplied by three is equal to 21. Weβll take the factor of negative one outside of the first term in our numerator. We then want to take the factor of 21 and multiply through our first set of parentheses. This gives us negative one multiplied by 63π₯ minus 147. And this is our final answer. Therefore, weβve shown that π¦ is equal to seven times the cot of three π₯ divided by three π₯ minus seven. Then dπ¦ by dπ₯ is equal to negative one time 63π₯ minus 147 times the csc squared of three π₯ minus 21 times the cot of three π₯ all divided by three π₯ minus seven squared.