Video Transcript
If 𝑥 squared plus 𝑥𝑦 plus three 𝑦 cubed equals 14, find d𝑦 by d𝑥 in terms of 𝑥 and 𝑦.
If we look at the equation of the curve we’ve been given, we see that it is given in terms of both 𝑥 and 𝑦. For example, we have this term here of plus 𝑥𝑦. And we have a term of plus three 𝑦 cubed. We’re also told that our derivative, d𝑦 by d𝑥, needs to be in terms of both 𝑥 and 𝑦. This tells us that we’re going to need to find d𝑦 by d𝑥 using implicit differentiation. The key principle behind implicit differentiation is the chain rule. Which tells us that if 𝑓 is a function of 𝑦 and 𝑦 is a function of 𝑓, 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. d𝑓 by d𝑥 equals by d𝑓 by d𝑦 times d𝑦 by d𝑥.
We’ll see how to apply the chain rule in order to differentiate implicitly in a moment. But, first, we just take the derivative of each side of this equation, with respect to 𝑥. We, then, note that the derivative of a sum can be written as the sum of the derivatives. So on the left-hand side, we can express this as the derivative, with respect to 𝑥, of 𝑥 squared plus the derivative, with respect to 𝑥, of 𝑥𝑦 plus the derivative, with respect to 𝑥, of three 𝑦 cubed.
Now, some of these terms are just constants or they’re in terms of 𝑥 only. In which case, their derivatives, with respect to 𝑥, can be found easily. For example, on the right-hand side, the derivative, with respect to 𝑥, of 14 is just zero, as 14 is a constant. And the derivative, with respect to 𝑥, of 𝑥 squared is just two 𝑥, if we recall the power rule of differentiation. But, what about the derivatives of the two middle terms, both of which involve 𝑦.
Well, this is where I will need to apply implicit differentiation. Let’s consider the derivative, with respect to 𝑥, of three 𝑦 cubed first. If we define the function 𝑓 of 𝑦 to be three 𝑦 cubed, then we recall that by the chain rule, 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. By applying the power rule of differentiation, we see that the derivative of 𝑓, with respect to 𝑦, is three times three 𝑦 squared which is nine 𝑦 squared. And, therefore, substituting into the chain rule, we have that the derivative of 𝑓, with respect to 𝑥, is equal to nine 𝑦 squared d𝑦 by d𝑥.
It’s absolutely fine that this derivative is also in terms of d𝑦 by d𝑥. We’ll see how to deal with this in a moment. Now, let’s consider how to differentiate the final term, the derivative, with respect to 𝑥, of 𝑥𝑦. Now, this term is in fact a product. It’s the product of 𝑥 and 𝑦. So we recall the product rule which tells us that if 𝑢 and 𝑣 are differentiable functions in 𝑥, then the derivative, with respect to 𝑥, of 𝑢𝑣 is equal to 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥. We can, therefore, let 𝑢 equal 𝑥 and 𝑣 equal 𝑦.
Before we can apply the product rule, we need to find each of their derivatives, the derivative of 𝑢, with respect to 𝑥, is just one. And in order to find the derivative of 𝑣, with respect to 𝑥, we need to apply the chain rule. The derivative of 𝑣, with respect to 𝑥, is equal to the derivative of 𝑣, with respect to 𝑦, multiplied by the derivative of 𝑦, with respect to 𝑥. 𝑣 is equal to 𝑦. So as the derivative of 𝑣, with respect to 𝑦, is the same as the derivative of 𝑦, with respect to 𝑦, which is just equal to one. But, multiplying by one has no effect. So the derivative of 𝑣, with respect to 𝑥, is just equal to the derivative of 𝑦, with respect to 𝑥, d𝑦 by d𝑥.
Now, we’re able to substitute into the product rule of differentiation. The derivative, with respect to 𝑥, of the product 𝑥𝑦 is equal to 𝑢 times d𝑣 by d𝑥. That’s 𝑥 times d𝑦 by d𝑥 plus 𝑣 times d𝑢 by d𝑥. That’s 𝑦 multiplied by one. So we have that the derivative, with respect to 𝑥, of 𝑥𝑦 is equal to 𝑥d𝑦 by d𝑥 plus 𝑦, which we can substitute back into our overall derivative.
Now, we have successfully differentiated the entire of each side of this equation with respect to 𝑥. We want to find d𝑦 by d𝑥. So we need to rearrange this equation to make d𝑦 by d𝑥 a subject. But, we note, first of all, that d𝑦 by d𝑥 appears in two places in this equation. So our first step is to move all of the terms that don’t involve d𝑦 by d𝑥 to the other side of the equation, giving 𝑥d𝑦 by d𝑥 plus nine 𝑦 squared d𝑦 by d𝑥 equals negative two 𝑥 minus 𝑦. As d𝑦 by d𝑥 appears in both terms on the left-hand side of our equation, the key step we need here is to factor d𝑦 by d𝑥 from both terms, giving 𝑥 plus nine 𝑦 squared multiplied by d𝑦 by d𝑥 equals negative two 𝑥 minus 𝑦.
We can then divide both sides of this equation by 𝑥 plus nine 𝑦 squared, to leave d𝑦 by d𝑥 on its own on the left-hand side. If we want, we can also factor a negative one from the numerator of the fraction to give d𝑦 by d𝑥 equals negative two 𝑥 plus 𝑦 over 𝑥 plus nine 𝑦 squared. So, by performing implicit differentiation and then rearranging the resulting equation to make d𝑦 by d𝑥 the subject, we found our expression for d𝑦 by d𝑥 in terms of 𝑥 and 𝑦. d𝑦 by d𝑥 is equal to negative two 𝑥 plus 𝑦 over 𝑥 plus nine 𝑦 squared.
