### Video Transcript

Differentiate the function π¦ equals three π to the power of π₯ minus five over the cube root of π₯.

The first thing weβre gonna do is actually rewrite our equation using exponent rules. The first rule weβre gonna use is that π₯ to the power of one over π is equal to the πth root of π₯, which means that we can actually rewrite our function as π¦ is equal to three π to the power of π₯ minus five over π₯ to the power of a third.

And then we can actually use a second exponent rule, which tells us that π₯ to the power of negative π is equal to one over π₯ to the power of π. So, therefore, again, we can rewrite our function. And now we can rewrite it as π¦ is equal to three π to the power of π₯ minus five π₯ to power of negative a third. And weβve got that point by applying each of our exponent rules that we talked about.

You could actually go straight from the original function to the final line that we have at the moment just by applying them both in one go. I just showed you how to do it step by step so you understand how theyβre used. Okay, great! So now letβs get on and differentiate our function.

So if weβre gonna find ππ¦ ππ₯, or weβre gonna find the differential of our function, the first term weβre gonna look at is this one. Our first term is three π to the power of π₯. And in this term, we can actually see that we have π to the power of π₯. And π is just a number where the slope of π to the power of π₯ is equal to π to the power of π₯. So therefore, if we differentiate our term, weβre gonna be left with three π to the power of π₯ because it doesnβt change. And this relationship can actually be proved using limits.

So now Iβm gonna move on to the second term. And weβre gonna get plus five over three. And we get that because if we get negative five, which is our coefficient, multiplied by our exponent, which is negative a third, weβre gonna get positive five over three. And then this is gonna be multiplied by π₯ to the power of negative four over three. And we get that because if we have negative one over three and then we subtract one or three over three, we get negative four over three.

So therefore, we can say that if we differentiate the function π¦ equals three π to the power of π₯ minus five over the cube root of π₯, weβre gonna get three π to the power of π₯ plus five over three π₯ to the power of negative four over three.