### Video Transcript

Differentiate β of π‘ equals five cube root of π‘ minus π to the power of π‘.

Now, in order to differentiate our function, the first thing we want to do is rewrite it. Weβre gonna rewrite it using exponents rather than the cube root. And we can use one of our exponent rules to help us do this. Because we know that if we have π₯ to any root, itβs gonna be equal to π₯ to the power of one over that root, so one over π in this case. So, in our case, the π is going to be three cause weβve got a cube root. So, we can rewrite our function as β of π‘ is equal to five π‘ to the power of a third minus π to the power of π‘.

So, now, what we have are two terms that we need to differentiate. So, we can now say that the derivative of β of π‘ is gonna be equal to five over three π‘ to the power of negative two over three. And we got that because what we have is the exponent multiplied by the coefficient, so a third multiplied by five, which gives a five over three. Then, what we do is we subtract one from the exponent, so weβve got a third minus three over three, which is the same as one. So, that gives us our negative two over three. So, weβre left with five over three π‘ to the power of negative two over three.

And then, thatβs the first term differentiated. So, if we move to the second term. Well, if we differentiate π to the power of π‘, weβre just gonna get π to the power of π‘. And the reason we get that is because if we differentiate π to power of π₯, then what we get is π to power of π₯ multiplied by the derivative of π₯. Or for instance if it was π to the power of ππ₯, itβd be π to the power of ππ₯ multiplied by the derivative of ππ₯. Well, we know that the derivative of π₯ is just one. So therefore, π to the power of π₯ multiplied by one is just π to the power of π₯, or in our case π to the power of π‘. So, great, weβve got it in this form.

But what we do want to do is rewrite it. And we can rewrite it by using some exponent laws. So, another of the exponent laws that we can use is the one that tells us that if weβve got π₯ to the power of negative π, so weβve got a negative exponent, then this is equal to one over π₯ to the power of π. So therefore, weβve got five over three and then weβve got π‘ to the power of two over three and then minus π to the power of π‘.

So then, we can use another experiment law that shows us that if weβve got π₯ to the power of π over π, this is equal to the π root of π₯ to the power of π. And therefore, if we apply this rule, we get five over three multiplied by the cube root of π‘ squared minus π to the power of π‘. And weβve done this because weβve applied the rule that was just seen because we had π‘ power of two over three. So, itβs the cube root and then π‘ squared. So, weβve now solved the problem because weβve said that the derivative of β of π‘ is equal to five multiplied by the cube root of π‘ minus π to power of π‘ is five over three multiplied by the cube root of π‘ squared minus π to the power of π‘.