Video: Differentiating Rational Functions Using the Quotient Rule

Find 𝑑𝑦/𝑑π‘₯, given that 𝑦 = (π‘₯Β³ + 7π‘₯Β² + 6)/(π‘₯ + 8).

04:47

Video Transcript

Find 𝑑𝑦 𝑑π‘₯, given that 𝑦 equals π‘₯ cubed plus seven π‘₯ squared plus six over π‘₯ plus eight.

Now, in order to actually solve this problem, what we’re gonna have to do is differentiate our function. But to enable us to do that, what we’re gonna use is the quotient rule. And we can use the quotient rule because our function is in the form 𝑒 over 𝑣. So as we’ve got that, what we’re gonna have a look at is what the quotient rule actually is. Well, the quotient rule tells us that 𝑑𝑦 𝑑π‘₯ is equal to 𝑣 𝑑𝑒 𝑑π‘₯ minus 𝑒 𝑑𝑣 𝑑π‘₯ over 𝑣 squared. So what this actually means in practice is 𝑣 multiplied by the derivative of 𝑒 minus 𝑒 multiplied by the derivative of 𝑣 all over 𝑣 squared.

Okay, great. So now, we’ve got the quotient rule. Let’s apply it to actually find 𝑑𝑦 𝑑π‘₯. So the first thing I’ve done is I’ve actually identified 𝑒 and 𝑣 in our question. So our 𝑒 is gonna be π‘₯ cubed plus seven π‘₯ squared plus six because that’s our numerator. And our 𝑣 is gonna be our denominator which is π‘₯ plus eight. So next, what I’m gonna do, I’m actually gonna find 𝑑𝑒 𝑑π‘₯. So I’m gonna differentiate our 𝑒 value we’ve got here. So we’ll differentiate π‘₯ cubed plus seven π‘₯ squared plus six.

Just a quickly recap, how we’re gonna differentiate each of these. What we’re gonna look at is the differentiations of general rules, so our power rule. And what it is, is if we’ve got a function that’s in the form π‘Žπ‘₯ to the power of 𝑏, then the first derivative of that function is gonna be equal to π‘Žπ‘π‘₯ to the power of 𝑏 minus one. So our coefficient multiplied by our exponent. And then, we reduce the exponent by one because it’s gonna be π‘Žπ‘π‘₯ to the power of 𝑏 minus one. Okay, we’ve just recapped that. Let’s move on and differentiate and find 𝑑𝑒 𝑑π‘₯.

So we go ahead and apply that power rule. And our first term is gonna be three π‘₯ squared. And that’s because our coefficient is one multiplied by the exponent three gives us three. And then we reduce the exponent by one, so π‘₯ to the power of two or π‘₯ squared. And our second term is just 14π‘₯, so positive 14π‘₯. The six, when we differentiate the six, it actually disappears cause if you differentiate just an integer on its own, we get zero. Okay, great. So we’ve now found 𝑑𝑒 𝑑π‘₯.

So now, we move on to 𝑑𝑣 𝑑π‘₯. So again, we differentiate π‘₯ plus eight. And when we do that, we just get one. And that’s because π‘₯ differentiates to one. And as before, the positive eight just differentiates to zero. Great, so we’ve now found 𝑑𝑒 𝑑π‘₯ and 𝑑𝑣 𝑑π‘₯. So now, we can move on to the final stage which is to actually apply the quotient rule to find out 𝑑𝑦 𝑑π‘₯.

So when we actually apply the quotient rule, we’re gonna get 𝑑𝑦 𝑑π‘₯ is equal to. And then first of all, 𝑣 𝑑𝑒 𝑑π‘₯. So our 𝑣, π‘₯ plus eight, multiplied by our 𝑑𝑒 𝑑π‘₯, which is three π‘₯ squared plus 14π‘₯. And then we’re gonna subtract π‘₯ cubed plus seven π‘₯ squared plus six because that’s our 𝑒 𝑑𝑣 𝑑π‘₯. And it’s just that value there because our 𝑑𝑣 𝑑π‘₯ is just one. Okay, great. And then finally, what we do is we actually divide by 𝑣 squared. So it’s π‘₯ plus eight squared. Okay, fabulous. We’re actually at the point now where we’ve put all of our values into our quotient rule. Let’s simplify to find 𝑑𝑦 𝑑π‘₯.

Well, our first stage is to actually expand the parentheses. So we’re gonna have π‘₯ multiplied by three π‘₯ squared which is gonna give us three π‘₯ cubed. And then, we have π‘₯ multiplied by positive 14π‘₯ which gives us plus 14π‘₯ squared. And then next, we have positive eight multiplied by three π‘₯ squared which gives us positive 24π‘₯ squared. So finally, we get plus 112π‘₯. And that’s because we had positive eight multiplied by positive 14π‘₯. Then we have minus π‘₯ cubed minus seven π‘₯ squared minus six.

So at this point, we actually can draw your attention to a possible common mistake. So as you can see, there’s a negative in front of the parentheses. I put the second part in a parentheses because actually just to remind you that actually everything inside there has to be negative. So it’s negative π‘₯ cubed minus seven π‘₯ squared minus six. Because commonly, people will just do minus π‘₯ cubed. But then, they’ll just do plus seven π‘₯ squared plus six. So be careful for that. Okay, and then this is all divided by π‘₯ plus eight squared.

And when we simplify this, we have three π‘₯ cubed minus π‘₯ cubed which gives us two π‘₯ cubed. Then we have 14π‘₯ squared plus 24π‘₯ squared minus seven π‘₯ squared which gives us positive 31π‘₯ squared. Then we have plus 112π‘₯ minus six over π‘₯ plus eight all squared.

So therefore, we can say that given that 𝑦 is equal to π‘₯ cubed plus seven π‘₯ squared plus six over π‘₯ plus eight, then 𝑑𝑦 𝑑π‘₯ is equal to two π‘₯ cubed plus 31π‘₯ squared plus 112π‘₯ minus six over π‘₯ plus eight all squared.

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