### Video Transcript

Differentiate π of π₯ is equal to π to the power of π₯ multiplied by the cos of π₯.

Weβre given a function π of π₯, and we need to differentiate this function. And since π is a function of π₯, this means we need to differentiate this with respect to π₯. To do this, letβs start by looking at our function π of π₯. We can see itβs the product of two functions. Itβs π to the power of π₯ multiplied by the cos of π₯. And we know we can find the derivative of the products of two functions by using the product rule. Letβs start by recalling the product rule.

The product rule tells us if we have a function π of π₯ which is the product of two differentiable functions π’ of π₯ times π£ of π₯, then π prime of π₯ is equal to π’ prime of π₯ times π£ of π₯ plus π£ prime of π₯ times π’ of π₯. And we can see this is whatβs happening in this question. We could set π’ of π₯ to be π to the power of π₯ and π£ of π₯ to be the cos of π₯. And we know both of these are differentiable. So we can find π prime of π₯ by using the product rule.

Now, to use the product rule, we see we need to find expressions for π’ prime of π₯ and π£ prime of π₯. Letβs start with π’ prime of π₯. Since π’ of π₯ is π to the power of π₯, π’ prime of π₯ will be the derivative of π to the power of π₯ with respect to π₯.

To evaluate this derivative, we need to recall the derivative of the exponential function π to the power of π₯ with respect to π₯ is equal to itself, π to the power of π₯. So by applying this, we just get π’ prime of π₯ is equal to π to the power of π₯.

We now need to find an expression for π£ prime of π₯. Since π£ of π₯ is the cos of π₯, this will be the derivative of the cos of π₯ with respect to π₯. To evaluate this, we need to recall one of our standard trigonometric derivative results. The derivative of the cos of π₯ with respect to π₯ is equal to negative the sin of π₯. So by applying this, we get that π£ prime of π₯ is equal to negative the sin of π₯. Weβre now ready to use the product rule to help us find an expression for π prime of π₯. Remember, itβs equal to π’ prime of π₯ times π£ of π₯ plus π£ prime of π₯ times π’ of π₯.

Substituting in our expressions for π’ of π₯, π£ of π₯, π’ prime of π₯, and π£ prime of π₯, we get that π prime of π₯ is equal to π to the power of π₯ times the cos of π₯ plus negative the sin of π₯ multiplied by π to the power of π₯. And we could leave our answer like this. However, we can also simplify by taking out the common factor of π to the power of π₯.

And this gives us our final answer: π to the power of π₯ times the cos of π₯ minus the sin of π₯. Therefore, given π of π₯ is equal to π to the power of π₯ multiplied by the cos of π₯, we were able to differentiate this with respect to π₯ by using the product rule. We got π prime of π₯ is equal to π to the power of π₯ times the cos of π₯ minus the sin of π₯.