# Question Video: Finding the First Derivative of the Quotient of Trigonometric and Linear Functions Using the Quotient Rule Mathematics • Higher Education

If π¦ = (tan 7π₯)/(9π₯), find dπ¦/dπ₯.

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### Video Transcript

If π¦ is equal to the tan of seven π₯ divided by nine π₯, find the derivative of π¦ with respect to π₯.

Weβre given π¦ as the quotient of two functions. Itβs the tan of seven π₯ divided by nine π₯. We need to use this to find the derivative of π¦ with respect to π₯. And thereβs a few different ways of doing this. For example, we could consider this as the tan of seven π₯ multiplied by one over nine π₯ and then use the product rule.

However, since π¦ is given as the quotient of two functions, weβll do this by using the quotient rule. First, we need to recall the quotient rule. It tells us the derivative of the quotient of two functions π’ over π£ is equal to π’ prime π£ minus π£ prime π’ all divided by π£ squared.

So to apply the quotient rule to π¦, we need to set π’ of π₯ to be the function in our numerator β thatβs the tan of seven π₯ β and π£ of π₯ to be the function in our denominator β thatβs nine π₯. We then need to find expressions for π’ prime and π£ prime.

Letβs start with π’ prime of π₯. Thatβs the derivative of the tan of seven π₯ with respect to π₯. And to evaluate this, we need to recall one of our standard trigonometric derivative results. We know for any real constant π, the derivative of the tan of ππ₯ with respect to π₯ is equal to π times the sec squared of ππ₯. In our case, the value of π is equal to seven. So using this, weβve shown π’ prime of π₯ is equal to seven times the sec squared of seven π₯.

We can also find an expression for π£ prime of π₯. Itβs the derivative of nine π₯ with respect to π₯. But nine π₯ is a linear function. So its slope will just be the coefficient of π₯, which is nine.

Weβre now ready to use the quotient rule to find an expression for dπ¦ by dπ₯. Itβs equal to π’ prime of π₯ times π£ of π₯ minus π£ prime of π₯ times π’ of π₯ all divided by π£ of π₯ all squared. Substituting in our expressions for π’ of π₯, π£ of π₯, π’ prime of π₯, and π£ prime of π₯, we get that dπ¦ by dπ₯ is equal to seven times the sec squared of seven π₯ multiplied by nine π₯ minus nine times the tan of seven π₯ all divided by nine π₯ all squared.

And now we can start simplifying. First, in our denominator, nine π₯ all squared can be simplified to give us 81π₯ squared. Next, we can cancel a shared factor of nine in our numerator and our denominator. And then, after some rearrangements, we get seven π₯ times the sec squared of seven π₯ minus the tan of seven π₯ all divided by nine π₯ squared. And this is our final answer.

Therefore, by using the quotient rule on π¦ is equal to the tan of seven π₯ all divided by nine π₯, we were able to show that dπ¦ by dπ₯ is equal to seven π₯ times the sec squared of seven π₯ minus the tan of seven π₯ all divided by nine π₯ squared.