### Video Transcript

Find the derivative of the function π½ of π is equal to the tan squared of π times π.

The question gives us a function π½ in terms of π. And it wants us to find the derivative of this function. Since our function is given in terms of π, we want to find the derivative of π½ with respect to π. So we want to find the derivative with respect to π of the tangent squared of ππ. Thereβs a few different ways of doing this. Weβre going to do this by realizing that the tan squared of ππ is just equal to the tan of ππ multiplied by the tan of ππ. Now, we can see the derivative that we need to evaluate is the derivative of the product of two functions. So we can do this by using the product rule. We recall the product rule tells us, for two functions π’ and π£, the derivative of π’ times π£ with respect to π is equal to π£ times dπ’ by dπ plus π’ times dπ£ by dπ.

So to apply this rule, weβll set both π’ and π£ equal to the tan of π times π. So to use the product rule, weβre going to need to find expressions for dπ’ by dπ and dπ£ by dπ. Both of these are equal to the derivative of the tan of ππ with respect to π. And we know, for any constant π, the derivative of the tan of π times π with respect to π is equal to π times the sec squared of π times π. And we can then use this to evaluate both of our derivatives. We get that dπ’ by dπ and dπ£ by dπ are both equal to π times the sec squared of π times π.

Weβre now ready to evaluate the derivative of the tan squared of π times π with respect to π by using the product rule. We get that itβs equal to π£ times dπ’ by dπ plus π’ times dπ£ by dπ. Substituting our expressions for π’, π£, dπ’ by dπ, and dπ£ by dπ, we get the dπ½ by dπ is equal to the tan of π times π multiplied by π times the sec squared of ππ plus the tan of ππ times π times the sec squared of ππ. And we can see that both of the expressions here which weβre adding together are equal.

So to add these two terms together, we can just multiply one of the terms by two. And we can rearrange this to give two π times the sec squared of ππ multiplied by the tan of ππ. And remember, this was an expression for the first derivative of our function π½ of π. So this is equal to π½ prime of π. Therefore, weβve shown that π½ of π is equal to the tan squared of ππ. Then π½ prime of π is equal to two π times the sec squared of ππ times the tan of ππ.