### Video Transcript

Find ππ¦ ππ₯, given that π¦ equals three π₯ to the power of four plus four π₯ squared plus six minus seven over π₯ to the power of seven minus eight over π₯ to the power of eight.

Now, in order to actually find ππ¦ ππ₯, what weβre gonna have to do is actually differentiate our function. However, before I do that, what I want to actually do is rewrite it in a way such that we remove the fractions. And to do this, what weβre gonna use is the exponent rule. And that rule tells us that if we have one over π to the power of π, then this is gonna be equal to π to the power of negative π.

Okay, great, so letβs use this and actually rewrite our function. So now, weβve actually used this. Weβve got π¦ is equal to three π₯ to the power of four plus four π₯ squared plus six minus seven π₯ to the power of negative seven minus eight π₯ to the power of negative eight. And the way in which we actually use our exponent rule to get these last two terms was by β if you wanted to think of it this way β we had seven multiplied by one over π₯ to the power of seven. So if I split it up like that, so therefore weβd actually get seven multiplied by π₯ to the power of negative seven. So we get seven π₯ to the power of negative seven.

Okay, now that we have this, we can actually get on and differentiate our function. Now in order to actually differentiate our function, we actually use something called the general power rule. And this tells us that if we have a function in the form ππ₯ to the power of π, then the derivative of this function is gonna be equal to πππ₯ to the power of π minus one. So what weβve actually done in practice is weβve multiplied the exponent by the coefficient β so ππ β and then weβve actually reduced the exponent by one β so π minus one.

Okay, so now what we can do is we can actually use this to differentiate our function. So then using a differentiation rule, we have our first term which is 12π₯ to the power of three. So the way we actually differentiated three π₯ to the power of four was we multiplied our coefficient by our exponent β so three by four β which gave us 12 and then itβs π₯ to the power of four minus one cause we subtract one from the exponent which gave us π₯ to the power of three. And then, our next term is plus eight π₯ and we got that in the same way we did four multiplied by two, coefficient multiplied by exponent, gives us eight. And then, we had π₯ to the power of two minus one gives us just π₯ or π₯ to the power of one.

Then, looking at our next term, this would just be plus zero which we wouldnβt usually write, but Iβm just putting in here so we can actually think about how we got that. Well, if we actually think about it like this, weβve got the derivative of six π₯ to the power of zero. And I say that because if we have six, then itβs just six multiplied by one. Well, π₯ to the power of zero is one.

So how would we actually differentiate that? Well, what weβd have is the coefficient of six multiplied by the exponent of zero and π₯ to the power of zero minus one. Well, zero multiplied by anything is just zero. So therefore, our term is going to be just zero.

Okay, great, letβs move on to the last two terms. Well, our next term is gonna be positive 49π₯ to the power of negative eight. Well, just to draw our attention to how we actually did this one just because itβs got a couple of negatives in there, weβve got the derivative of negative seven π₯ to the power of negative seven. Well, then we have a negative seven because thatβs our coefficient multiplied by negative seven which is our exponent. So therefore, negative multiplied by negative gives us a positive and thatβs 49. And then, we have π₯ to the power of negative seven minus one, which gives us negative eight.

And then, our final term works in exactly the same way and we get positive 64π₯ to the minus nine because we had negative eight multiplied by negative eight which gives us positive 64. And then, we had π₯ to the power of negative eight minus one which gives us π₯ to the power of negative nine. Okay, fantastic, weβve actually got each of our terms. Now, letβs actually simplify and put it back into its original form.

So therefore, we can say that given that π¦ equals three π₯ to the power of four plus four π₯ squared plus six minus seven over π₯ to the power of seven minus eight over π₯ to the power of eight, then ππ¦ ππ₯ is gonna be equal to 12π₯ cubed plus eight π₯ plus 49 over π₯ to the power of eight plus 64 over π₯ to the power of nine. And we got the last two terms in that form just by using our exponent rule that we used at the beginning, which was that one over π to the power of π is equal to π to the power of negative π.