Practical lesson 3. Differentiation of a composite function and differentiation of implicitly defined function
The theme of this practical lesson is differentiation of a composite function and differentiation of implicitly defined function.We will discuss the function of several variables.
Let's move on to solving problems.
Find all the partial derivatives of a composite function (see. Video). z is a function of the variables u and v, where each of these variables is a function of variables x and y.
Draw a diagram corresponding to this situation (see the video.): Z depends on the variables u and v, each of these variables depend on x and y.
These transitions correspond to partial derivatives: dz / du, du / dx, if we differentiate in x, we can write u’x (this is the same), dz / dv, dv / dx.
Then the value of partial derivative dz / dx (see the video.), dz / du * du / dx is the first transition of the upper arrows, dz / dv * dv / dx is the second transition through variable v.
Similarly, the value of the partial derivative of y corresponds to differentiation and will occur only in the last phase of y (see. Video).
To write out the answer, we need to know the value of all partial derivatives z’u, z’v and the partial derivatives of function u of x and y and of function v of x and y.
Let's find all these partial derivatives: dz/du=1+v2, dz/dv= 2uv.
The value of partial derivatives of function u: du / dx = 1 / (x-y), du / dy - it's the same derivative, just multiplied by -1, we write 1 / (y-x), so that we have fewer digits.
For the derivative of function v: dv / dx, x is a variable in the base, then it is a power function, we get: y*xy-1; dv/dy, this is the exponential function, this is xy*ln x.
Let us write out only values of the partial derivative dz / dx.
What do we see?
The value dz/dx = (1+v) * 1/(x-y) + 2uvyxy-1.
Circlewthat we have received, and analyze the situation: dz / dy is absolutely similar, all the values are written out.
Look what we got: partial derivative z’x already contains four variables, it is not very good.
What's the matter?
How to improve the situation: it is best to add "where u = ln (x-u), v = xy».
In reality, we got again a complex function of two variables x and y.
We proceed to the next problem (see. Video).
Look what we see: f is, generally speaking, a function of two variables, here they are not introduced, and we will write the label, because how else can we differentiate?
The easiest way is to take and to say that the problem can be formulated slightly differently: z = f (u, v), where u = 2x-1 and v= sin2x, it is possible to write the system sign, it can be separated by comma, in general, it is not necessary.
Let's draw a diagram: z depends on u and v, which in turn depend on a single variable x.
Let's write down here with strokes (see the video.): z’u, z’v or if we have function f, we can write f’u, f’v at the same point.
And here is the transition, look (see the video.): u and v are functions of one variable x, so there appears u' and v'.
There is no partial derivatives.
If we used letters d, we would have written «u’=du/dx», we would have used these.letters d and
What we see: z ', if this is the function of variable x, let's write it like this: f’u*u’+f’v*v’.
What do we see?
Let's write the final answer: z’= f’u*u’+f’v*v’= f’u(u, v)*2+f’v(u, v)*v’, where v’= 2sin(x)cos(x)=sin(2x).
Still, at the beginning the problem didn’t mention any u and v, so for the answer to comply with the problem, we will write "where u=2*x-1, v=sin2(x)».
The problem is solved.
Now we will talk about and solve problems on the function given implicitly, on the differentiation of such functions.
Functions and in both cases (see. Video), we see this is one and the same. It is not possible to express z in terms of x and y, so dependence of z on x and y is implicit dependence.
Let's check that the point (2, 1, 0) satisfies.
If we substitute 2 in place of x, 1 instead of y, 0 instead of z, we get: 1 + 2 = 3 – it is true that the set of space points is not empty, there is some surface, a lot of points in space.
Let's do it one by one.
The first task: find the value or rather the partial derivatives of the implicitly defined function.
The first of these problems (see. Video).
First, we remember that when we differentiate we bring the equation to this form (see. Video), so we'll move everything to the left side and say that the function of three variables look like this (see. Video).
It remains to remember the formula: z’x= -F’x/F’z, and z’y= -F’y/F’z t these are formulas of partial value function defined implicitly.
All we need to know is the partial derivatives of function F, I will not rewrite the variables with respect to each of the variables x, y, z. F’x,: in this case, F is a function of three variables, and x, y, z in this recording are completely equivalent, each of the variables is independent.
In differentiation with respect to x, the first summand becomes a constant, and y is also a constant, so it is y on x. we look at F’y, only the third summand contains a variable y and the partial derivative is equal to x. F’z = ez – 1.
It remains to apply the formula: z’x - F’x is divided by F’z and the minus sign not to write a minus sign before the fraction, we will immediately send it to th denominator, and this is 1-ez (see the video.) z’y is x divided by the same value, and minus sign and this is: x/(1-ez).
The problem is solved.
Let's move on to the next problem.
We need to find the equation of the tangent plane to the surface (see Fig. Video).
Let us remember how to determined that.
We say that this is point P0.
The value of the partial derivative of function F at the point P0 is F’x(P0), F’x(P0)( x-x0) + F’y(P0)(y-y0) + F’z(P0)(z-z0) = 0.
Coordinates x0, y0, z0 are known, and the value of the partial derivatives are to be found.
For what function did we just do that: for function F (x, y, z) this is еz-z+xy-3.
We just found out that we will have to find the value of the partial derivative with respect to x, at point P0,.
We still have to return to the problem: it is xon y, y is on x and z is еz – 1.
We will use these are the values F’x(P0) = y, F’y(P0) = x, F’z(P0) = ez - 1, in principle, we could write immediately, it is not difficult to find.
Here, we will find the value of derivatives at the point P0, then we substitute coordinates (2, 1, 0): y = 1, then F’x(P0) = 1; at point P0 x=2, F’y(P0) = 2; z=0, and then we calculate and get F’z(P0) = 0.
It remains to write the equation of the plane.
The first coefficient is 1, so there remains only (x-x0), x0 is 2 and we get (x-2), the first summand is recorded, + 2 * (y-1), the derivative of y = 2, y0 = 1 and the last summand will be equal to 0, so do not write anything and equate to 0.
Open the brackets: x + 2y-4 = 0.
The equation of the tangent plane is obtained.