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

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

Given that 4π₯Β² β 5π₯ + 8π¦Β² β 3π¦ = 1, find ππ¦/ππ₯.

03:29

### Video Transcript

Given that four π₯ squared minus five π₯ plus eight π¦ squared minus three π¦ equals one, find ππ¦ by ππ₯.

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

As this equation is a relation between the variables π₯ and π¦, 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 times π₯ to the π is π times π times π₯ to the π minus one. We find that our first derivative is eight π₯.

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

So what is π by ππ₯ of eight π¦ squared? We differentiate eight π¦ squared with respect to π¦ and get 16π¦ and then you multiply by ππ¦ by ππ₯. Similarly, for π by ππ₯ of three π¦, you differentiate three π¦ with respect to π¦ to get three and then multiply by ππ¦ by ππ₯. 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₯, π¦, and ππ¦ by ππ₯ and we can rearrange this equation to make ππ¦ by ππ₯ the subject β that is we can write ππ¦ by ππ₯ in terms of π₯ and π¦. The first thing to do here is to factor out ππ¦ by ππ₯ 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 ππ¦ by ππ₯ on the left. And dividing through by this something, we get that ππ¦ by ππ₯ is negative eight π₯ plus five over 16π¦ minus three.

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

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 itβs best to leave the answer in terms of both π₯ and π¦.

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