Hi everyone, welcome to Mathematically Inclined. I am Neha, your Math Mentor. And today we are discussing derivative of implicit functions.
This is also part 3 of differentiation. If you haven't seen the previous videos, then I'll be leaving the link in the description box below. Also, for the videos on other chapters, I have created separate playlists. The link for the same is in the description box. So, let's get started.
Let's try and find out what are implicit functions. Here, I have two examples in front of me. Both of them are functions in terms of x and y. So, if I check out the first function, let me try and express one variable in terms of other.
By try expressing Let's say y in terms of x. After taking y common from these two terms, it could be conveniently done this way. That means y becomes a function of x.
Let me try a similar exercise with this one. So if I try to take let's say y common from here, I end up getting y as this but y is not entirely in terms of x. Similarly, if I try expressing x in terms of y, we land into a similar situation.
A function of the form fxy is said to be implicit if it is not possible to express y entirely in terms of x or vice versa. So, the problem is how do you differentiate such functions? Well, let's see that using an example.
Now suppose you are given this function to be differentiated with respect to x. You know this is an implicit function because you cannot express one in terms of another. So we differentiate with respect to x.
If you look at the first term, it is the product of two functions. So derivative of x which is 1 into the second function as it is plus x as it is into derivative of y which is again 1 but this time since you are differentiating with respect to x and the variable is y, therefore you get dy upon dx according to the chain rule. Similarly, y square gives us the derivative 2y but once again since you are differentiating with respect to x and the variable is y, so it becomes dy into dx. Moving on to the right side, derivative of tan x with respect to x is simply secant square x.
Plus derivative of y is 1 into once again dy by dx. Now we have clubbed all the terms containing dy by dx, brought them to one side and the rest of the terms I am taking to the right side. So secant square x minus y.
Your final dy by dx in this case is this. Wasn't that simple? So the trick is do not try to express one function in terms of other variable.
Simply differentiate it and use chain rule wherever required. Now looking at this second example, once again the situation is similar. You have to evaluate gy by dx.
So do not even try to fix the terms. You can directly differentiate. To know sine square y is the same as sine y whole square. It is of the form derivative of x square which would be 2x. So here the derivative of sine y whole square would be 2 sine y.
But we notice that the function is sine y. So you differentiate this sine y, you get cos y. Now you notice the function is y whereas we wanted to differentiate with respect to x.
So further you get into dy by dx. This is cos of xy. So cos has a derivative minus sin of xy.
But when you move ahead you realize the function was xy. So you need to differentiate that. It is once again the product of two functions.
So Derivative of x which is just 1 into y as it is plus x as it is into derivative of y 1 into dy by dx. And the right side is pi which being a constant gives us 0. So we repeat the process of clubbing the terms with dy by dx now. 2 sin y cos y becomes sin 2 y.
And if I open the brackets the term containing dy by dx is taken with it. And the remaining term is minus sin xy which I push to the right side to make it sin of xy. Therefore...
dy by dx becomes sin of xy whole upon sin 2y minus x sin xy. So moving ahead with the third question, once again you differentiate with respect to x. Look at the first term. So once again we need to apply the product rule which gives us First derivative of x cube is 3x square into y square as it is plus first term x cube as it is into derivative of y square which would be 2y. And now by this time you would have understood we further multiply this with dy by dx is equal to.
Now if you remember from the previous video. Log of x plus y, the derivative would be 1 upon x plus y. But once again, x plus y was not a standard function.
So, you need to differentiate this further. And we get derivative of x is simply 1 and derivative of y is 1 into dy by dx. We get 1 plus dy by dx plus derivative of sine is cos.
e raised to power x and further differentiating e to the power x with respect to x again gives us e to the power x. Please note here it is already x so you do not have to differentiate it further or even if you do you are just getting one. However had it been y you would have gotten an extra dy by dx. You get gy by gx and you take the terms containing dy by dx And the rest of the terms on the right side. Now the process is the same.
Your dy by dx is your right side divided with this entire term. We have another question here. Let's try to tackle it using implicit differentiation.
By which I mean differentiate it directly. So if I just take up this equation and I differentiate with respect to x. So derivative of the first term is dy by dx into second function as it is plus y as it is into derivative of under root 1 minus x square which gives us 1 upon 2 1 minus x square raised to power half minus 1 into further you will have to apply your chain rule this gives us into minus 2x.
plus derivative of x is just 1 into second function as it is, plus x as it is into when you repeat the process of under root 1 minus y square, you get 1 by 2, 1 minus y square raised to power half minus 1 into derivative of y square would be minus 2y, but You are differentiating with respect to x. So, further into dy by dx. When you differentiate the right side which is just 1, you get equal to 0. Now, simplifying this further, you get dy by dx into under root 1 minus x square.
These 2 and 2, they get cancelled and you have minus xy upon. This becomes 1 minus x square raised to power minus half which comes down as. under root 1 minus x square plus under root 1 minus y square and after cancelling 2 again you get minus xy upon under root 1 minus y square into dy by dx equal to 0. Now let's sort the terms containing dy by dx. So for that these two terms would be clubbed and this implies dy by dx Taken common, you get under root 1 minus x square minus xy upon under root 1 minus y square is equal to, let's take these two terms to the right.
You would be getting xy upon under root 1 minus x square minus under root 1 minus y square. So now on taking the LCM on this side, you get under root 1 minus y square in the denominator and this expression in the numerator. Likewise on the right side, when you take under root 1 minus x square as the LCM, you get a very similar looking term.
If you take minus sign. common from either your LHS or your RHS, you would realize that your numerator is absolutely the same. So therefore, after canceling these two and taking minus sign common, your dy by dx would be under root of 1 minus y square upon 1 minus x square. I hope you enjoyed learning the concept of implicit functions.
and its derivatives. If yes, then give this video a big thumbs up, subscribe right away and also comment because your feedback is really important. I'll see you with the next video. Until then, bye-bye.