Video: Differentiating Rational Functions Using the Quotient Rule

Find the first derivative of function 𝑦 = 4𝑥/(9𝑥² − 7).

03:40

Video Transcript

Find the first derivative of the function 𝑦 equals four 𝑥 over nine 𝑥 squared minus seven.

Now note to actually solve this problem and find our first derivative, what we’re gonna use is the quotient rule. And we can use the quotient rule because our function is actually in the form 𝑢 over 𝑣 because it’s actually has a fraction. And the quotient rule tells us that d𝑦 d𝑥 — so our first derivative — is equal to 𝑣 d𝑢 d𝑥 minus 𝑢 d𝑣 d𝑥 over 𝑣 squared. So, in practical terms, what is means is, actually, our denominator multiplied by our numerator’s derivative minus our numerator multiplied by our denominator’s derivative and this is all divided by the denominator squared.

Okay, so now that we have this, let’s start to use it and actually find out what the derivative of our function is. Well, the first stage is actually to identify what our 𝑢 and 𝑣 are. So our 𝑢 is gonna be four 𝑥 and our 𝑣 is nine 𝑥 squared minus seven and that’s because they’re in fact our numerator and denominator, respectively. So next, what we’re gonna need to is actually differentiate four 𝑥 and differentiate nine 𝑥 squared minus seven.

I’m gonna start with 𝑢. So we’re gonna have d𝑢 d𝑥 is equal to- well, just to remind us how we’re actually gonna differentiate. If we know that we got a function in the form 𝑎𝑥 to the power of 𝑏, then we know that our first derivative is gonna be 𝑎𝑏 𝑥 to the power of 𝑏 minus one. So what we’ve done is we multiplied the coefficient by the exponent and then we’ve actually reduced one from the exponent. So that’s why I’ve got 𝑎𝑏 𝑥 to the power of 𝑏 minus one.

Okay, so let’s apply this and actually differentiate 𝑢. So if we differentiate 𝑢, we’d just gonna get four and that’s because we multiplied our coefficient which is four by the exponent which is just one so that’s four. And then, we had 𝑥 to the power of- well, one minus one is zero, so 𝑥 to the power of zero is just one. So we’re just left with four.

Okay so now, let’s move on to d𝑣 d𝑥. So d𝑣 d𝑥 is just gonna give us 18𝑥 and that’s because we’ve differentiated the first term — so nine multiplied by two is 18 — and reduced the exponent by one. So we just get 𝑥 on its own. And this is just it because actually negative seven differentiates to zero because if you have an integer, they all differentiate just to zero. Okay, we now have d𝑢 d𝑥 and d𝑣 d𝑥, Let’s move on and actually apply the quotient rule.

So first of all, what we’ve got is 𝑣 d𝑢 d𝑥. So it’s gonna be four multiplied by nine 𝑥 squared minus seven or run it the other way round just because it’s an easy one, we’re gonna expand the parenthesis. And this is minus four 𝑥 multiplied by 18𝑥 and this is because this is our 𝑢 d𝑣 d𝑥 and then this is all over 𝑣 squared — so nine 𝑥 squared minus seven all squared.

So now, we’re gonna move on to our next stage, which is actually expanding the parentheses. So then, we have d𝑦 d𝑥 is equal to and then our first term on numerator is 36𝑥 squared because four multiplied by nine 𝑥 squared and then we have minus 28 because we have four multiplied by negative seven. And then, we finally have minus 72𝑥 squared. And this is because we have four 𝑥 multiplied by 18𝑥 which gives us 72𝑥 squared and then we have the negative sign in the middle. Okay, great, and this is all divided by our 𝑣 squared which is nine 𝑥 minus seven all squared.

So then, we get on to our final stage, which is actually to simplify the numerator. And all we need to do is actually subtract 72𝑥 squared from 36𝑥 squared which is negative 36𝑥 squared.

So we can now say that the first derivative with the function 𝑦 equals four 𝑥 over nine 𝑥 squared minus seven is equal to negative 36𝑥 squared minus 28 over nine 𝑥 minus seven all squared.

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