### Video Transcript

Differentiate π¦ equals root π₯ multiplied by five π₯ root π₯ minus four π₯ over root π₯ minus one.

So, the first thing weβre gonna do before we differentiate is rewrite our function in exponent form. And to do that, weβre gonna use a couple of exponent laws. So, weβve got π₯ to the power of a half is equal to root π₯. π₯ to the power of negative π is equal to one over π₯ to the power of π. And when we do that, weβre gonna get π¦ is equal to π₯ to the power of a half multiplied by five π₯ multiplied by π₯ to the power of a half minus four π₯ multiplied by π₯ to the power of negative a half minus one.

So, now to simplify it further, what weβre going to do is weβre gonna use another exponent law. And that one is that if we have π₯ to the power of π multiplied by π₯ to the power of π, then itβs equal to π₯ to the power of π plus π. So, we add the exponents.

So, then we got π¦ is equal to π₯ to the power of a half multiplied by five π₯ to the power of three over two. Thatβs because we added the exponents. So, we had one add a half, so it gives us one and a half, or three over two, minus four π₯ to the power of a half minus one. And we got the four π₯ to the power of a half because we had four π₯ multiplied by π₯ to the power of negative a half. So, thatβs adding the exponents. So, one add negative a half, which is the same as one take away a half, which is just a half.

Now the final stage for simplifying before we differentiate is to expand the parentheses. So, we have π₯ to the power of a half multiplied by five π₯ to the power of three over two. So, when we multiply these together, weβre gonna get five π₯ squared. Thatβs because if we add the exponents, weβll have a half plus three over two, or three-halves, which gives us four-halves, which is the same as two.

And then the second term is going to be four π₯. Thatβs cause we have π₯ to the power of a half multiplied by four π₯ to the power of a half. So, again, we add the exponents. A half plus a half is one. And then, finally, we have π₯ to the power of a half. So, great! We now have it in a form thatβs gonna be easy to differentiate.

So, we can now say that the derivative of our function, so π¦ prime, or can be written as ππ¦ ππ₯, itβs gonna be equal to 10π₯ for our first term. Thatβs cause we have the exponent multiplied by the coefficients, so two multiplied by five, which is 10. And then we reduce the exponent by one. So, we go from two to one, so we get 10π₯. And then, we get minus four. Thatβs because, in the same way, we differentiate four π₯. So, we multiply the exponent which is one by four, which gives us four, then reduce the exponent to zero. So, weβre just left with four. Because π₯ to the power of zero is just one.

And finally, the last term is negative a half π₯ to the power of negative a half. And again, thatβs because we had a half multiplied by one, which gives us a half. Then we reduce the exponent by one, so we go from a half to negative a half.

So, now finally, weβre just gonna rewrite our answer back in its original form. And weβre gonna do that by using our exponent laws to rewrite the final term. So therefore, we can say that the derivative of π¦ equals root π₯ multiplied by five π₯ root π₯ minus four π₯ over root π₯ minus one is going to be 10π₯ minus four minus one over two root π₯. And we got that because we had π₯ to the power of negative a half. Well, we know that if we have π₯ to the power of a half, itβs gonna give us root π₯. But then we use the law that tells us that if we have π₯ to the power of negative π, itβs one over π₯ to the power π. So, thatβs how we write our final answer.