### Video Transcript

Given that π¦ is equal to four times the tan of nine π₯ over five, determine π¦ double prime.

Weβre given π¦ is a function in π₯. In fact, itβs a trigonometric function. And weβre asked to determine an expression for π¦ double prime. Remember, the prime symbol means we need to differentiate. So π¦ double prime means we need to find the second derivative. And since this is a function of π₯, we need to differentiate twice with respect to π₯.

So the first thing we need to do is find an expression for π¦ prime. Thatβs the first derivative of π¦ with respect to π₯. We can see we need to differentiate the tangent function. And we know this is a standard trigonometric derivative result we should commit to memory.

We know for any real constant π, the derivative of the tan of ππ₯ with respect to π₯ is equal to π times the sec squared of ππ₯. So we can find an expression for π¦ prime by setting our value of π equal to nine over five. And it might be easier to see this by rewriting nine π₯ over five as nine-fifths times π₯. So by setting π equal to nine-fifths and applying our derivative rule, we get π¦ prime is equal to four times nine-fifths multiplied by the sec squared of nine π₯ over five. And we can simplify this slightly. Four multiplied by nine is equal to 36.

So weβve found an expression for π¦ prime. But we need to find π¦ double prime. This means we need to differentiate π¦ with respect to π₯. So π¦ double prime will be equal to the derivative of 36 over five times the sec squared of nine π₯ over five with respect to π₯. And thereβs a lot of different ways of evaluating this derivative. For example, we could use the product rule. We could apply our trigonometric identities and use the quotient rule. We could also use the chain rule. However, weβre going to use the general power rule.

We recall the general power rule tells us for differentiable function π of π₯ and real constant π, the derivative of π of π₯ all raised to the πth power with respect to π₯ is equal to π times π prime of π₯ multiplied by π of π₯ raised to the power of π minus one. We could evaluate this directly. However, weβll start by taking the constant factor of 36 over five outside of our derivative. So now weβre just differentiating the sec squared of nine π₯ over five. Weβll set our value of the exponent π equal to two. And weβll set our inner function π of π₯ to be the sec of nine π₯ over five. And now we can see this is in exactly a form in which we can apply the general power rule.

However, before we can apply the general power rule, we need to find an expression for π prime of π₯. Since π of π₯ is a trigonometric function, weβre going to need to use one of our standard trigonometric derivative results. We recall for any real constant π, the derivative of the sec of ππ₯ with respect to π₯ is equal to π sec of ππ₯ multiplied by the tan of ππ₯. And for our function π of π₯, we can see the value of π is equal to nine over five. So by setting π equal to nine over five, we get π prime of π₯ is equal to nine-fifths times the sec of nine π₯ over five multiplied by the tan of nine π₯ over five.

And now that we found an expression for π prime of π₯, we can substitute this and π is equal to two into our general power rule to find an expression for π¦ double prime. We get that itβs equal to 36 over five times two multiplied by nine-fifths sec of nine π₯ over five times the tan of nine π₯ over five multiplied by the sec of nine π₯ over five raised to the power of two minus one. And of course we can simplify this. Weβll start by calculating our coefficient. We get that itβs equal to 648 divided by 25. Next, in our exponent, we have two minus one. This simplifies to give us one. And raising a function to the first power doesnβt change the function. So we can rewrite π¦ double prime as the following expression.

And we can see we also have the sec of nine π₯ over five multiplied by the sec of nine π₯ over five. We can rewrite this as the sec squared of nine π₯ over five. And this gives us our final answer. Therefore, we were able to show if π¦ is equal to four tan of nine π₯ over five, then π¦ double prime will be equal to 648 over 25 times the sec squared of nine π₯ over five multiplied by the tan of nine π₯ over five.