### Video Transcript

Find the first derivative of the function π¦ equals nine π₯ plus five π₯ multiplied by four π₯ squared plus five over π₯ all squared.

The first step with this question is to write it out with both the parentheses in there. Thatβs cause we have four π₯ squared plus five over π₯ all squared. So, Iβve written it out so that we have nine π₯ plus five π₯ multiplied by four π₯ squared plus five over π₯ multiplied by four π₯ squared plus five over π₯.

And now what Iβm gonna do is Iβm gonna rewrite five over π₯ in exponent form. And Iβm gonna do that using the rule that π₯ to the power of negative one is equal to one over π₯. So, now in each of the parentheses, we have five π₯ to the power of negative one as the last term. So, now what Iβm gonna do is expand the parentheses. And to expand the parentheses, Iβm gonna use another exponent rule. And that is that if we have π₯ to the power of π multiplied by π₯ to the power of π, we add the exponents. So, itβs π₯ to the power of π plus π.

So therefore, the first termβs gonna be 16π₯ to the power of four. And thatβs cause four multiplied by four is 16. And then π₯ squared multiplied by π₯ squared gives us π₯ to the power of four, as two add two is equal to four. And then weβre gonna get add 20π₯. Thatβs cause four π₯ squared multiplied by five π₯ to the power of negative one. Four multiplied by five is 20.

And then, we got π₯ squared multiplied by π₯ to the power of negative one. So, two add negative one is just one. So, we just have π₯. And then, we have plus another 20π₯ cause, again, we have positive five π₯ to the power negative one multiplied by four π₯ squared. And then, finally, we add 25π₯ to the power of negative two. And thatβs cause we have five π₯ to the power of negative one multiplied by five π₯ to the power of negative one.

So, now the next step is to collect the like terms. Weβve got two π₯ terms here, so we can collect them. So, when weβve done that, we got nine π₯ plus five π₯ multiplied by 16π₯ to the power of four plus 40π₯. And thatβs cause we have 20π₯ plus 20π₯ plus 25π₯ to the power of negative two. So, now what we need to do is expand the parentheses by multiplying five π₯ by each of the terms inside.

So, the first term is gonna be 80π₯ to the power of five. So, weβve got nine π₯ plus 80π₯ to the power of five. And thatβs cause we have five π₯ multiplied by 16π₯ to the power of four. Then, plus 200π₯ squared. And then, finally, plus 125π₯ to the power of negative one. And thatβs because if we have five π₯ multiplied by 25π₯ to the power of negative two. But if weβve got π₯ on its own, thatβs the same as π₯ to the power of one. So, weβve got one add negative two, which gives us the power of negative one, or the exponent of negative one.

So, now we rewrite our function in descending exponents of π₯. And weβve got it in a form now thatβs easy to differentiate. So, in the first derivative, the first termβs gonna be 400π₯ to the power of four. And just to remind us how we differentiate, we multiply the exponent by the coefficient, so five by 80, which gives us our 400. And then, we reduce the exponent by one. So, we get five minus one, which gives us four. So, we got 400π₯ to the power of four.

Then, plus 400π₯ then plus nine cause if we differentiate nine π₯, we just get nine. And then, finally, minus a 125π₯ to the power of negative two. And itβs minus this because weβre gonna multiply by the exponent, which is a negative one. So, if we multiply 125 by negative one, we get negative 125.

So therefore, we can say that the first derivative of the function π¦ equals nine π₯ plus five π₯ multiplied by four π₯ squared plus five over π₯ all squared is going to be 400π₯ to the power of four plus 400π₯ minus a 125π₯ to the power of negative two plus nine.