### Video Transcript

Find the derivative of the function π of π‘ is equal to π to the power of ππ‘ times the sin of ππ‘.

The question gives us a function π in terms of π‘. It wants us to find the derivative of this function. So, we want to differentiate this with respect to π‘. We can see that our function π of π‘ is the product of two functions. Itβs the product of an exponential function and a trigonometric function. And we know how to differentiate the product of two functions by using the product rule.

If we have that π of π‘ is equal to the product of two functions π’ of π‘ and π£ of π‘, then the product rule tells us π prime of π‘ is equal to π£ of π‘ times π’ prime of π‘ plus π’ of π‘ times π£ prime of π‘. So, to use the product rule, weβll set our function π’ of π‘ to be π to the power of ππ‘, and weβll set our function π£ of π‘ to be the sin of ππ‘. To use the product rule, weβre going to need to find the expressions for π’ prime of π‘ and π£ prime of π‘.

Letβs start with π’ prime of π‘. It is the derivative of π to the power of ππ‘ with respect to π‘. And we know how to evaluate this derivative. We know for any constant π, the derivative of π to the power of ππ‘ with respect to π‘ is π times π to the power of ππ‘. So, in our case, we just have π is equal to π. So, we get that π’ prime of π‘ is π times π to the power of ππ‘.

We now need to find an expression for π£ prime of π‘. Thatβs the derivative of the sin of ππ‘ with respect to π‘. And this is a standard trigonometric derivative result. We know for any constant π, the derivative of the sin of ππ‘ with respect to π‘ is equal to π times the cos of ππ‘. So, using this result with π equal to π, we get that π£ prime of π‘ is equal to π times the cos of ππ‘.

Weβre now ready to find an expression for our derivative of π. By the product rule, we have π prime of π‘ is equal to π£ of π‘ times π’ prime of π‘ plus π’ of ~~π£~~ [π‘] times π£ prime of π‘. Substituting in our expressions for π’, π£, π’ prime, and π£ prime, we get that π prime of π‘ is equal to the sin of ππ‘ times ππ to the power of ππ‘ plus π to the power of ππ‘ times π cos of ππ‘. And we can simplify this expression. We will take out our common factor of π to the power of ππ‘. Doing this and rearranging, we get π to the power of ππ‘ times π sin of ππ‘ plus π cos of ππ‘. And this is our final answer.

Therefore, weβve shown if π of π‘ is equal to π to the power of ππ‘ times the sin of ππ‘, then π prime of π‘ is equal to π to the power of ππ‘ times π sin of ππ‘ plus π cos of ππ‘.