Video: Differentiating Rational Functions Using the Quotient Rule

Differentiate 𝑓(π‘₯) = (4π‘₯Β² βˆ’ 5π‘₯ + 8)/(3π‘₯ βˆ’ 4).

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Video Transcript

Differentiate 𝑓π‘₯ equals four π‘₯ squared minus five π‘₯ plus eight over three π‘₯ minus four.

So in order to actually differentiate our function, what we’re gonna use is something called the quotient rule. And we know that we’re gonna use the quotient rule because actually our function is in the form 𝑒 over 𝑣, because it’s actually a fraction.

What the quotient rule actually tells us is that 𝑑𝑦 𝑑π‘₯, so our derivative, is equal to 𝑣 𝑑𝑒 𝑑π‘₯ minus 𝑒 𝑑𝑣 𝑑π‘₯ over 𝑣 squared. What that actually means in practice is its our 𝑣 multiplied by our derivative of 𝑒 and then it’s minus 𝑒 multiplied by our derivative of 𝑣 and then all divided by 𝑣 squared. Okay, fab! Now that we have this, let’s use it and actually find out what the derivative of our function actually is.

So my first step is to actually identify our 𝑒 and our 𝑣. So I use our numerator, so four π‘₯ squared minus five π‘₯ plus eight. And our 𝑣 is going to be three π‘₯ minus four, which is our denominator. So the first thing we’re gonna do is actually differentiate 𝑒. So we differentiate four π‘₯ squared minus five π‘₯ plus eight.

And just to quickly remind us how we do that, if we’ve got 𝑓π‘₯ in the form π‘Žπ‘₯ to the power of 𝑏, then the derivative of our function, or could be known as 𝑑𝑦 𝑑π‘₯, is equal to π‘Žπ‘, so coefficient multiplied by the exponent, and then π‘₯ to the power of 𝑏 minus one. So you subtract one from the exponent.

Okay, so now let’s use this and differentiate 𝑒. So we’re gonna get eight π‘₯ minus five. And just to kind of remind us how we did that, the first one was gonna be eight π‘₯, because we’ve got four multiplied by two, an exponent, multiplied by coefficient, which gives us eight, and then π‘₯ to the power of one, because you actually subtract one from the exponent. So two minus one is one. So we’re left with eight π‘₯.

Okay, great! So now let’s move on to 𝑑𝑣 𝑑π‘₯. Well, if we differentiate 𝑣, so we’re gonna find 𝑑𝑣 𝑑π‘₯. And this is just gonna be equal to three. And that’s because if we differentiate three π‘₯, we get three. And if we differentiate just an integer on its own, you get zero.

Okay, so great! We’ve now found 𝑑𝑒 𝑑π‘₯ and 𝑑𝑣 𝑑π‘₯. So we can now actually apply our quotient rule to find 𝑑𝑦 𝑑π‘₯. So first of all, we’re gonna have 𝑣 𝑑𝑒 𝑑π‘₯. So it’s gonna be three π‘₯ minus four multiplied by eight π‘₯ minus five. And then we’re gonna have minus three multiplied by four π‘₯ squared minus five π‘₯ plus eight. So that’s our 𝑒 𝑑𝑣 𝑑π‘₯. And I’ve put it the other way round just because it’s easier when we’re gonna expand the parentheses in the next stage. And then this is all divided by three π‘₯ minus four all squared. So it’s our 𝑣 squared.

Okay, so now let’s move on to the next stage, which I’ve already said is gonna be expanding the parentheses on the numerator. So first of all, I’ve got three π‘₯ multiplied by eight π‘₯, which will give me 24π‘₯ squared. Then three π‘₯ multiplied by negative five gives us negative 15π‘₯, then minus 32π‘₯ because we had negative four multiplied by eight π‘₯, and then finally plus 20 because we had negative four multiplied by negative five. And a negative multiplied by negative gives us positive.

Then we’ve got minus 12π‘₯ squared, because we’ve got negative three multiplied by four π‘₯ squared, plus 15π‘₯, negative three multiplied by negative five π‘₯, again negative multiplied by negative, and then finally minus 24. Okay, then this is all divided by three π‘₯ minus four all squared.

So we now are gonna move on to our final stage, which is actually to simplify our numerator. So then we’re gonna have our first term. It’s gonna be 12π‘₯ squared. And that’s because we’ve got 24π‘₯ squared minus 12π‘₯ squared. Then we get minus 32π‘₯. And that’s because we had negative 15π‘₯ minus 32π‘₯ plus 15π‘₯. So therefore, the 15π‘₯s cancel out. We have minus 32π‘₯.

And then, finally, we get minus four. And this is because we had positive 20 minus 24. And then this is all over three π‘₯ minus four all squared. So great! We can say that if we differentiate the function four π‘₯ squared minus five π‘₯ plus eight over three π‘₯ minus four, then 𝑑𝑦 𝑑π‘₯ is gonna be equal to 12π‘₯ squared minus 32π‘₯ minus four over three π‘₯ minus four all squared.

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