Video Transcript
If three 𝑥 cubed minus three 𝑥𝑦 minus five 𝑦 equals two, what is the value of d𝑦 by d𝑥 when 𝑥 equals negative two?
So we’ve been asked to find the value of the first derivative of this function when 𝑥 takes a particular value. Looking at the function though, we can see that it has been defined implicitly. Instead of being given in the form 𝑦 equals, we see that 𝑦 is embedded within the definition of the function. And it appears in two places.
In order to find this derivative then, we’re going to need to use the method of implicit differentiation. We’re going to take the derivative of each side of this equation with respect to 𝑥. Now, on the left-hand side, we have the derivative of a sum or in fact a difference. And we recall that if we’re taking the derivative of a sum or difference, this is equal to the sum or difference of the derivatives. We can essentially differentiate each term separately and then add or subtract their derivatives. So we have that the derivative with respect to 𝑥 of three 𝑥 cubed minus the derivative with respect to 𝑥 of three 𝑥𝑦 minus the derivative with respect to 𝑥 of five 𝑦 is equal to the derivative with respect to 𝑥 of two.
Now some of these derivatives are straightforward to find. For example, on the right-hand side, we know that the derivative with respect to 𝑥 of a constant is just zero. Similarly, for the first term, the derivative with respect to 𝑥 of three 𝑥 cubed, we can apply the power rule of differentiation. As we’re finding the derivative with respect to 𝑥 of a function in 𝑥 only. And we see that the derivative with respect to 𝑥 of three 𝑥 cubed is equal to three multiplied by three 𝑥 squared. That’s just nine 𝑥 squared. But we’re then trying to find the derivatives with respect to 𝑥 of two terms that also involve 𝑦. And this is where we need to recall implicit differentiation.
Now implicit differentiation is just an application of the chain rule. It tells us that if 𝑓 is a function of 𝑦 and 𝑦 is a function of 𝑥. Then the derivative of 𝑓 with respect to 𝑥 is equal to the derivative of 𝑓 with respect to 𝑦 multiplied by the derivative of 𝑦 with respect to 𝑥. It’s actually going to be more straightforward to apply this to the second of the two terms we’re looking at, the derivative with respect to 𝑥 of five 𝑦. So let’s see what this will look like.
Here, our function 𝑓 is the function five 𝑦. The derivative of 𝑓 with respect to 𝑦, first of all, we can find using the power rule of differentiation. And it’s just equal to five. We then multiply this by the derivative of 𝑦 with respect to 𝑥. So we find that the derivative with respect to 𝑥 of five 𝑦 is five d𝑦 by d𝑥. So we’re nearly there. But we need to find the derivative with respect to 𝑥 of three 𝑥𝑦. And this is a little more complicated because it is a product. We can think of three 𝑥𝑦 as the product of three 𝑥 and 𝑦. And so we’re also going to need to apply the product rule.
The product rule tells us that, for two differentiable functions 𝑢 and 𝑣, the derivative with respect to 𝑥 of their product 𝑢𝑣 is equal to 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥. We’ll therefore let 𝑢 equal three 𝑥 and 𝑣 equal 𝑦. The derivative of 𝑢 with respect to 𝑥 is no problem. It’s just three by applying the power rule of differentiation. For the derivative of 𝑣 with respect to 𝑥, we’ll need to apply implicit differentiation again. Here our function 𝑓 is actually 𝑣. And it’s just 𝑦. So by using implicit differentiation, first we differentiate 𝑣 with respect to 𝑦. The derivative of 𝑦 with respect to 𝑦 is just one. And then we multiply by the derivative of 𝑦 with respect to 𝑥. Of course, multiplying by one has no effect. So we see that d𝑣 by d𝑥 is just equal to d𝑦 by d𝑥.
Now we can substitute into the product rule. The derivative with respect to 𝑥 of 𝑢𝑣 or three 𝑥𝑦 is equal to 𝑢 times d𝑣 by d𝑥. That’s three 𝑥 times d𝑦 by d𝑥. We then add 𝑣. That’s 𝑦 times d𝑢 by d𝑥, which is three. And so we find that the derivative with respect to 𝑥 of three 𝑥𝑦 is three 𝑥 d𝑦 by d𝑥 plus three 𝑦. Remember though that we were subtracting this derivative. So each of those terms will be negative in our original equation. We’ve now completed our differentiation of each side of this equation. Giving nine 𝑥 squared minus three 𝑥 d𝑦 by d𝑥 minus three 𝑦 minus five d𝑦 by d𝑥 is equal to zero.
Remember, we’re looking to find the value of d𝑦 by d𝑥 when 𝑥 is equal to negative two. So our next step is to rearrange this equation to make d𝑦 by d𝑥 the subject. We start by collecting all of the terms involving d𝑦 by d𝑥 on one side of the equation. And all the terms that don’t involve d𝑦 by d𝑥 on the other. Giving three 𝑥 d𝑦 by d𝑥 plus five d𝑦 by d𝑥 equals nine 𝑥 squared minus three 𝑦. The next key step we need is to factor the left-hand side of the equation. d𝑦 by d𝑥 appears in two places. But by factoring d𝑦 by d𝑥 from each term, we can express the left-hand side as a product. Three 𝑥 plus five multiplied by d𝑦 by d𝑥.
To complete our rearrangement, we need to divide by that factor of three 𝑥 plus five, leaving d𝑦 by d𝑥 on its own on the left-hand side and nine 𝑥 squared minus three 𝑦 over three 𝑥 plus five on the right-hand side. We now have an expression for d𝑦 by d𝑥 in terms of both 𝑥 and 𝑦. Now we’re asked to find the value of d𝑦 by d𝑥 for particular 𝑥-value when 𝑥 is equal to negative two. We can substitute negative two for 𝑥 into our expression for d𝑦 by d𝑥. But we also need to find the value of 𝑦 when 𝑥 is equal to negative two, because our derivative is in terms of both 𝑥 and 𝑦.
To find the value of 𝑦 when 𝑥 is equal to negative two, we substitute negative two for 𝑥 in the original equation. This will simplify to negative 24 plus six 𝑦 minus five 𝑦 equals two. Which will further simplify by adding 24 to each side and subtracting the 𝑦 terms to 𝑦 equals 26. We now know that when 𝑥 is equal to negative two, 𝑦 is equal to 26. And so we have a pair of 𝑥𝑦-values that we can substitute into our derivative. So substituting negative two for 𝑥 and 26 for 𝑦 gives d𝑦 by d𝑥 equals nine multiplied by negative two squared minus three multiplied by 26 over three multiplied by negative two plus five. That’s nine times four, which is 36, minus 78 over negative six plus five. 36 minus 78 is negative 42. And negative six plus five is negative one. So we have negative 42 over negative one, which is just 42. So by applying implicit differentiation and the product rule, we found that the value of d𝑦 by d𝑥 when 𝑥 is equal to negative two is 42.
Now, it is just worth pointing out that, in this particular question, it would actually have been possible to rearrange the original equation to make 𝑦 the subject and hence specify the equation in an explicit form. Doing so, would have given an equation in the form 𝑦 equals a quotient. And we could’ve then applied the quotient rule to find an expression for d𝑦 by d𝑥 in terms of 𝑥 only. Whilst that would be a perfectly acceptable alternative method for this question, it’s unusual for that to be possible in a question of this type. And so it’s far better to make sure you’re comfortable with the process of implicit differentiation.
