Question Video: Finding and Evaluating the First Derivative of Polynomial Functions Using the Chain Rule and the Product Rule | Nagwa Question Video: Finding and Evaluating the First Derivative of Polynomial Functions Using the Chain Rule and the Product Rule | Nagwa

Question Video: Finding and Evaluating the First Derivative of Polynomial Functions Using the Chain Rule and the Product Rule Mathematics

Find the first derivative of the function 𝑦 = π‘₯⁴ (4π‘₯ + 9)⁹ at π‘₯ = βˆ’2.

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Video Transcript

Find the first derivative of the function 𝑦 equals π‘₯ to the power of four multiplied by four π‘₯ plus nine to the power of nine at π‘₯ equals negative two.

Well, the first thing we need to do to actually solve this problem is actually differentiate our function. Well, we can see that our function is in the form 𝑦 equals 𝑒𝑣. So therefore, we’re actually gonna use the product rule to help us differentiate it. And the product rule states that 𝑑𝑦 𝑑π‘₯ is equal to 𝑒 𝑑𝑣 𝑑π‘₯ plus 𝑣 𝑑𝑒 𝑑π‘₯.

So first of all, we pick out what’s gonna be our 𝑒 and our 𝑣. So our 𝑒 is going to be π‘₯ to the power of four and our 𝑣 is equal to four π‘₯ plus nine to the power of nine. So now, what we want to do is actually find out what 𝑑𝑒 𝑑π‘₯ and 𝑑𝑣 𝑑π‘₯ actually are. So I’m gonna start with 𝑑𝑒 𝑑π‘₯ and 𝑑𝑒 𝑑π‘₯ is gonna be equal to four π‘₯ cubed. And that’s because we’ve obviously differentiated our π‘₯ to the power of four.

And just to remind us how we actually did that, what we did is we actually multiplied the coefficient of our term by the exponent β€” so four multiplied by one. And then what we did is we actually subtracted one from our exponent β€” so four minus one. So that gave us four π‘₯ to the power of three.

Okay, great, so now let’s move on and differentiate 𝑣 to find 𝑑𝑣 𝑑π‘₯. Well, now to find 𝑑𝑣 𝑑π‘₯, what we need to do is we actually need to use a rule to help us actually differentiate four π‘₯ plus nine to the power of nine. And the rule we’re actually gonna use is the chain rule. And the chain rule actually tells us that 𝑑𝑦 𝑑π‘₯ is equal to 𝑑𝑦 𝑑𝑒 multiplied by 𝑑𝑒 𝑑π‘₯.

Okay, so now we have this rule, let’s look at our term and see if we can apply it. Well, first of all, we’re gonna have a look at 𝑒 and say that it’s equal to four π‘₯ plus nine. So therefore, 𝑑𝑒 𝑑π‘₯ is just gonna be four because if you differentiate four π‘₯, you get four and if you differentiate nine, you just get zero.

Okay, so now we can move on to 𝑦. Well, 𝑦 would be equal to 𝑒 to the power of nine. So therefore, 𝑑𝑦 𝑑𝑒 is equal to nine 𝑒 to the power of eight. So then if we apply the chain rule, we can say that 𝑑𝑦 𝑑π‘₯ is equal to four multiplied by nine 𝑒 to the power of eight, which is equal to 36𝑒 to the power of eight. So then, we just substitute back in that 𝑒 is equal to four π‘₯ plus nine and we get that 𝑑𝑦 𝑑π‘₯ is equal to 36 multiplied by four π‘₯ plus nine to the power of eight. So therefore, we can say the 𝑑𝑣 𝑑π‘₯ is equal to 36 multiplied by four π‘₯ plus nine to the power of eight.

So therefore, now that we found our 𝑑𝑣 𝑑π‘₯ and our 𝑑𝑒 𝑑π‘₯, what we can actually do is apply our product rule to find 𝑑𝑦 𝑑π‘₯. So the first derivative of our function π‘₯ to the power of four multiplied by four π‘₯ plus nine to the power of nine. So therefore, we get 𝑑𝑦 𝑑π‘₯ is equal to π‘₯ to the power of four because that’s our 𝑒 and then multiplied by our 𝑑𝑣 𝑑π‘₯ which is 36 multiplied by four π‘₯ plus nine to the power of eight plus our 𝑣 which is four π‘₯ plus nine to the power of nine multiplied by our 𝑑𝑒 𝑑π‘₯ which is four π‘₯ cubed.

So great, we’re actually at the stage where we found the first derivative of our function. But what do we do now? So now, what we need to do is actually look at what the first derivative is going to be when π‘₯ is equal to negative two. And in order to do this, we need to substitute in π‘₯ is equal to negative two into our first derivative.

So we’re gonna get that the first derivative with negative two substituted in for π‘₯ is equal to negative two to the power of four multiplied by 36 times four times negative two plus nine to the power of eight plus four multiplied by negative two plus nine to the power of nine multiplied by four multiplied by negative two cubed which is gonna be equal to 16 multiplied by 36 plus four multiplied by negative eight which is equal to 544.

So therefore, we can say that the first derivative of the function 𝑦 equals π‘₯ to the power of four multiplied by four π‘₯ plus nine to the power of nine at π‘₯ equals negative two is equal to 544.

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