Question Video: Differentiating Polynomial Functions | Nagwa Question Video: Differentiating Polynomial Functions | Nagwa

# Question Video: Differentiating Polynomial Functions Mathematics • Second Year of Secondary School

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Determine the first derivative of the function ๐ฆ = 2๐ฅ(9๐ฅยฒ โ 3๐ฅ) + 10๐ฅ.

03:21

### Video Transcript

Determine the first derivative of the function ๐ฆ equals two ๐ฅ multiplied by nine ๐ฅ squared minus three ๐ฅ plus 10๐ฅ.

So now, the first stage we actually need to complete is to expand our parentheses. So in order to do this, what weโre gonna do is multiply out two ๐ฅ by each term. So weโre gonna start with two ๐ฅ multiplied by nine ๐ฅ squared which will give us 18๐ฅ cubed. And then, weโre gonna have two ๐ฅ multiplied by negative three ๐ฅ. Very important here to make sure that we do include the sign. So negative, so itโs two ๐ฅ multiplied by negative three ๐ฅ. So therefore, our next term is gonna be minus six ๐ฅ squared. And finally, we add on our 10๐ฅ.

Okay, great. So we now have the function ๐ฆ equals 18๐ฅ cubed minus six ๐ฅ squared plus 10๐ฅ. So now that weโve actually expanded the parentheses, what we can do is actually differentiate each of our terms individually. So before we do this, letโs remind ourselves how we actually differentiate.

So if we look at differentiation and weโve got a function in the form ๐๐ฅ to the power of ๐, then the derivative of this function is gonna be equal to ๐๐๐ฅ to the power of ๐ minus one. So what weโve done is weโve multiplied the exponent by the coefficient. And then what weโve done is reduce the exponent itself by one. So we get ๐๐๐ฅ to the power of ๐ minus one. Okay, great. So now we kind of recapped this. Letโs use it to actually differentiate each of our terms.

So Iโm gonna go through them each step by step, just so we can see exactly what each of our terms is going to be. So first of all, if we differentiate 18๐ฅ cubed, this is gonna be equal to 18 multiplied by three. So the coefficient multiplied by the exponent. And then ๐ฅ to the power of three minus one which would give us 54๐ฅ squared. Okay, great. We found our first term in the derivative. Itโs also worth noting at this point, Iโve put our first derivative as ๐๐ฆ ๐๐ฅ. But with the notation, you could also see it as ๐ฆ prime or ๐ ๐ฅ prime. They both mean the same thing. It means the first derivative.

Okay, great. So now, letโs move on to our second term. So for our second term, weโre gonna differentiate negative six ๐ฅ squared. Iโve kept the negative in because actually itโs the sign thatโs relevant to this term. However, it wonโt actually affect the differentiation itself. And this is gonna give us negative six multiplied by two because thatโs how again our coefficient multiplied by our exponent. And then ๐ฅ to the power of two minus one because we reduce one from our exponent which will give us negative 12๐ฅ. Okay, great. We found our second term.

So therefore, we can now move on to our last term. Well, our last term is just gonna give us plus 10. And then just to remind ourselves how we actually got that, weโve got 10 multiplied by one because weโve got our coefficient 10. And then the exponent will be one, if itโs ๐ฅ on its own. And then ๐ฅ to the power of one minus one. And if you think about that, that will give us ๐ฅ to the power of zero. And we know anything to the power of zero is just equal to one. So therefore, weโre just left with 10.

So therefore, we can say that the first derivative of the function ๐ฆ equals two ๐ฅ multiplied by nine ๐ฅ squared minus three ๐ฅ plus 10๐ฅ is equal to 54๐ฅ squared minus 12๐ฅ plus 10.

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