Question Video: Finding the First Derivative of a Function Defined Implicitly Using Implicit Differentiation

Given that 4π‘₯Β² βˆ’ 5π‘₯ + 8𝑦² βˆ’ 3𝑦 = 1, find 𝑑𝑦/𝑑π‘₯.

03:29

Video Transcript

Given that four π‘₯ squared minus five π‘₯ plus eight 𝑦 squared minus three 𝑦 equals one, find 𝑑𝑦 by 𝑑π‘₯.

We could try to rearrange this equation so that we had 𝑦 in terms of π‘₯ and then differentiate in the normal way. This is tricky because not only do we have this minus three 𝑦 term, but also this eight 𝑦 squared. And so we have a quadratic in 𝑦. It’s possible to solve this quadratic and find 𝑦 in terms of π‘₯ and then differentiate. But there is a better way.

As this equation is a relation between the variables π‘₯ and 𝑦, we can differentiate both sides with respect to π‘₯. We can differentiate the left-hand side term by term, using the fact that the derivative with respect to π‘₯ of π‘Ž times π‘₯ to the 𝑛 is π‘Ž times 𝑛 times π‘₯ to the 𝑛 minus one. We find that our first derivative is eight π‘₯.

Similarly, 𝑑 by 𝑑π‘₯ of five π‘₯ is five. This term is slightly more difficult to differentiate. How do you find 𝑑 by 𝑑π‘₯ of a function of 𝑦? Here, we can use the chain rule: 𝑑𝑓 by 𝑑π‘₯ is 𝑑𝑓 by 𝑑𝑦 times 𝑑𝑦 by 𝑑π‘₯. And so just using slightly different notation here, we see that to differentiate a function of 𝑦 with respect to π‘₯, we differentiate that function of 𝑦 with respect to 𝑦 and then we just multiply by 𝑑𝑦 by 𝑑π‘₯.

So what is 𝑑 by 𝑑π‘₯ of eight 𝑦 squared? We differentiate eight 𝑦 squared with respect to 𝑦 and get 16𝑦 and then you multiply by 𝑑𝑦 by 𝑑π‘₯. Similarly, for 𝑑 by 𝑑π‘₯ of three 𝑦, you differentiate three 𝑦 with respect to 𝑦 to get three and then multiply by 𝑑𝑦 by 𝑑π‘₯. The right-hand side is more straightforward. So the derivative of one with respect to π‘₯ is just zero.

We now have an equation in terms of π‘₯, 𝑦, and 𝑑𝑦 by 𝑑π‘₯ and we can rearrange this equation to make 𝑑𝑦 by 𝑑π‘₯ the subject β€” that is we can write 𝑑𝑦 by 𝑑π‘₯ in terms of π‘₯ and 𝑦. The first thing to do here is to factor out 𝑑𝑦 by 𝑑π‘₯ in the two terms where it appears. Then, we can subtract eight π‘₯ minus five from both sides and then we just have something times 𝑑𝑦 by 𝑑π‘₯ on the left. And dividing through by this something, we get that 𝑑𝑦 by 𝑑π‘₯ is negative eight π‘₯ plus five over 16𝑦 minus three.

This is our final answer. Note that it is written in terms of both π‘₯ and 𝑦. You might feel slightly uneasy about this if you’re accustomed to writing 𝑑𝑦 by 𝑑π‘₯ in terms of π‘₯ alone. But to write 𝑑𝑦 by 𝑑π‘₯ in terms of π‘₯ alone, we’d have to write 𝑦 in terms of π‘₯. And as we discussed at the start of the video, that’s a bit tricky. And not only that, we’d probably find that the expression for 𝑦 in terms of π‘₯ is quite ugly. And so we get quite an ugly expression for 𝑑𝑦 by 𝑑π‘₯.

So for this question and for questions involving implicit differentiation more generally β€” that is when you differentiate both sides with respect to a variable β€” it’s best to leave the answer in terms of both π‘₯ and 𝑦.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.