Let the function of two changes be given. We give the argument an increase, but the argument is too much invariable. The same function removes the increase, as it is called a private increase for the change and is assigned:

Similarly, fixing the argument and giving the increment to the argument, we take away the private increase in the function behind the change:

The value is called the greatest increase in the function in points.

Appointment 4. The private function of the two changeable ones is called between the change of the private increase of the function until the change of the given change, if the remainder of the zero (i.e. the boundary) remains. It is signified privately like this: either, or.

In this rank, for the appointed mayor:

Private func- tions are calculated according to the very rules and formulas, as if the function is one of change, it is protected from its own, that it is differentiating according to change, it is important to be constant, and when differentiating it according to change, it should be

Example 3. Know private fun functions:

Solution. a) In order to know the important constant value of that differential as a function of one variable:

Similarly, with respect to the constant value, we know:

Appointment 5. The total differential of a function is the sum of creations of private similar functions on the increase of independent independent ones, tobto.

Looking back at the fact that the differentials of independent changes are growing with their increments, that is. , the formula for the total differential can be written in the form

Example 4. Calculate the final differential of the function.

Solution. Oskіlki behind the formula of the total differential is known

Private holidays of the highest order

Private holidays are called private holidays of the first order or first private holidays.

Appointments 6. Private functions of a different order are called private functions of the first order.

Private chotiri of a different order. Vons are designated as follows:

Similarly, private losses of the 3rd, 4th and higher orders are assigned. For example, for the function may:

Private holidays of a different order, taken from different changes, are called changed private holidays. For function є pokhіdnі. It’s respectful that you’re in a mood, if you’re fluent without interruption, there’s room for jealousy.

Example 5. Change private functions in a different order

Solution. Private first order functions found in application 3:

Differentiation and change x and y, otrimaemo

Virishuvati physics problems or apply mathematics is absolutely impossible without knowledge about that method of calculation. Pokhіdna is one of the most important to understand mathematical analysis. We decided to dedicate this fundamental topic to today's article. What is so bad, what kind of physical and geometrical change, how to spoil a good function? All meals can be taken in one: how can I understand how to go?

Geometrical and physical sense similar

Come on, function f(x) , is given in the singing interval (a,b) . The points x and x0 lie up to the th interval. When changing x, the function itself changes. Change of argument - difference of yoga value x-x0 . What difference is recorded as delta x and is called the greater argument. Changing or increasing the function is called the difference in the value of the function at two points. Appointment of travel:

Pokhіdna funktsії y point - between the increase of the function at tsіy point to the zbіlshennya argument, if the remainder is zero.

Otherwise, you can write it like this:

What is the sense of such a boundary? And the axis is yaki:

similar to the function at the point, the tangent of the kuta between the points OX is similar to the graph of the function at the tsij point.

Physical sense of the day: pokhіdna paths for an hour dorovnyuє shvidkostі rectilinear ruhu.

Definitely, we can see from school hours that swedishness is a private road. x=f(t) that hour t . Average speed for one hour:

Schob to recognize the security of the rush at the moment of the hour t0 it is necessary to calculate between:

First rule: blame the constant

The constant can be blamed for the bad sign. More than that - it requires work. When vyrishenny applied mathematics take as a rule - how can you ask viraz, obov'azkovo ask .

butt. Let's calculate the cost:

Rule to friend: Pokhіdna sumi funktsіy

Pokhіdna sumi dvoh funktsіy dorivnyuє sumі pokhіdnih tsikh funktsіy. The same is true for similar retail functions.

It does not suggest a proof of the theorem, but rather a practical example.

Know related functions:

Rule of three: bad work of functions

Pokhіdna create two functions that are differentiated, calculated by the formula:

Example: know the following functions:

Solution:

Here it is important to say about the number of folding similar functions. Pokhіdna foldable function is more expensive to supplement the pokhіdnoї tsієї funktsії behind the intermediate argument to the worse of the intermediate argument behind the independent change.

At the point of view, the application of mi zustrіchaєmo viraz:

In this case, the intermediate argument is 8x for the fifth step. In order to calculate the cost of such a virase, it is important to calculate the value of the outer function for the intermediate argument, and then multiply by the value of the non-intermediate argument for the independent change.

Rule four: similar to private two functions

The formula for choosing a similar part of two functions:

We tried to tell you about the holidays for teapots from scratch. This topic is not so simple, as it turns out, it’s possible: the butts often have pasta on the butts, so be careful when counting the ones.

For some reason, for other topics, you can turn to the student service. For a short term, we will help you to compose the checklist and sort out the tasks, so we didn’t deal with the calculation of the last ones earlier.

Let's look at the function in two ways:

Shards of change $x$ and $y$ are independent, for such a function it is possible to provide an understanding of private information:

Private function $f$ at point $M=\left(((x)_(0));((y)_(0)) \right)$ for change $x$ -

\[(((f)")_(x))=\underset(\Delta x\to 0)(\mathop(\lim ))\,\frac(f\left(((x)_(0) )+\Delta x;((y)_(0)) \right))(\Delta x)\]

In the same way, you can assign a private fee for a change of $y$:

\[(((f)")_(y))=\underset(\Delta y\to 0)(\mathop(\lim ))\,\frac(f\left(((x)_(0) );((y)_(0))+\Delta y \right))(\Delta y)\]

In other words, in order to know the private func- tions of some of the change, it is necessary to fix the decision of the change, the crime of shukano, and then we will know the zvichayna to look after the price of the change.

Sounds like the main trick for counting such lousy ones: just take into account that everything is changing, krym tsієї, є constant, after which differentiate the function so that you would differentiate the “singular” - from one zminnoy. For example:

$\begin(align)& ((\left(((x)^(2))+10xy \right))_(x))^(\prime )=((\left(((x)^(2 ) )) \right))^(\prime ))_(x)+10y\cdot ((\left(x \right))^(\prime ))_(x)=2x+10y, \\& ( ( \left(((x)^(2))+10xy \right))_(y))^(\prime )=((\left(((x)^(2)) \right))^( \ prime ))_(y)+10x\cdot ((\left(y \right))^(\prime ))_(y)=0+10x=10x. \\\end(align)$

It is obvious that it is normal to give private holidays from different changes. Why is it more important to understand, why, let's say, in the first one we were calmly charged $10y$ s-pid of a bad sign, and in the other - the first one was zeroed out. Everything is conceived through those that all letters, krіm zminnoi, for some kind of differentiation, are respected by constants: they can be blamed, spat, etc.

What is "private fun"?

Today we will talk about the functions of a few changers and about private holidays in them. First of all, what is the function of a few replacements? Dosi mi called to change the function like $y\left(x \right)$ or $t\left(x \right)$, otherwise change that one-single function in it. Now there will be only one function in us, and there will be a change of sprat. If you change $y$ and $x$ the value of the function will change. For example, if $x$ increases twice, the value of the function changes, if $x$ changes, but $y$ does not change, the value of the function changes itself.

It was understood that the function in the form of a number of variables, just like in one of the variables, can be differentiated. However, the oskіlki zmіnnykh kіlka, then it is possible to differentiate from different zmіnnyh. For whom, specific rules are blamed, which are the same when differentiating one change.

First for everything, if we want to lose our functions, if we are somehow changeable, then we are to blame, for what kind of change we are supposed to leave - that’s why it’s called a private mess. For example, we have a function of two different ones, and we can fix її like $x$, so $y$ are two private ones that are similar to the skin of the zminnyh.

In a different way, if we have fixed one of the zminnykh and we start to respect privately after it, then everything else that enters to the function function is respected by constants. For example, $z\left(xy \right)$, as we are important to privately go over $x$, then, squinting, demi-simply $y$, we are important to be a constant and to be treated by itself as a constant. Zokrema, when counting bad things, we can blame $y$ for the shackle (we have a constant), but when counting bad money, as we have here, it’s like a virus to avenge $y$ and not avenge $x$, then it’s good virazu dorivnyuvatime "zero" like a good constant.

At first glance, you can get away that I tell you about it in a folded way, and a lot of learners stray on the cob. There is nothing supernatural among the private ones, and we are changing from the butt of specific tasks.

Responsible for radicals and rich members

Manager No. 1

Sob not to waste an hour, from the very cob we’ll start with serious butts.

For starters, I guess the following formula:

This is the standard table value, as we know from the standard course.

It's good for someone to use $z$ like this:

\[(((z)")_(x))=((\left(\sqrt(\frac(y)(x)) \right))^(\prime ))_(x)=\frac( 1)(2\sqrt(\frac(y)(x)))((\left(\frac(y)(x) \right))^(\prime ))_(x)\]

Let's once again, the shards under the roots cost not $x$, but some other viraz, in this case $\frac(y)(x)$, then we speed up the standard tabular values, and then, the shards under the roots cost not $x $, and another viraz, it is necessary for us to multiply our expenses for one more viraz for the other viraz. Let's start stepping on the cob:

\[((\left(\frac(y)(x) \right))^(\prime ))_(x)=\frac((((((y)"))_(x))\cdot xy \cdot ((((x)"))_(x)))(((x)^(2)))=\frac(0\cdot xy\cdot 1)(((x)^(2) ) )=-\frac(y)(((x)^(2)))\]

Let's turn to our virazu and write down:

\[(((z)")_(x))=((\left(\sqrt(\frac(y)(x)) \right))^(\prime ))_(x)=\frac( 1)(2\sqrt(\frac(y)(x)))((\left(\frac(y)(x) \right))^(\prime ))_(x)=\frac(1) (2\sqrt(\frac(y)(x)))\cdot \left(-\frac(y)(((x)^(2))) \right)\]

Everything is in principle. However, it is wrong to leave її in such a look: it’s not handy to beat such a construction for the distant ones, so let’s do it in a trifle:

\[\frac(1)(2\sqrt(\frac(y)(x)))\cdot \left(-\frac(y)(((x)^(2))) \right)=\frac (1)(2)\cdot \sqrt(\frac(x)(y))\cdot \frac(y)(((x)^(2)))=\]

\[=-\frac(1)(2)\cdot \sqrt(\frac(x)(y))\cdot \sqrt(\frac(((y)^(2)))(((x)^ (4))))=-\frac(1)(2)\sqrt(\frac(x\cdot ((y)^(2)))(y\cdot ((x)^(4)))) =-\frac(1)(2)\sqrt(\frac(y)(((x)^(3))))\]

Vidpovid found. Now let's deal with $y$:

\[(((z)")_(y))=((\left(\sqrt(\frac(y)(x)) \right))^(\prime ))_(y)=\frac( 1)(2\sqrt(\frac(y)(x)))\cdot ((\left(\frac(y)(x) \right))^(\prime ))_(y)\]

Vipishemo okremo:

\[((\left(\frac(y)(x) \right))^(\prime ))_(y)=\frac((((((y)"))_(y))\cdot xy \cdot ((((x)"))_(y)))(((x)^(2)))=\frac(1\cdot xy\cdot 0)(((x)^(2) ) )=\frac(1)(x)\]

Now we write:

\[(((z)")_(y))=((\left(\sqrt(\frac(y)(x)) \right))^(\prime ))_(y)=\frac( 1)(2\sqrt(\frac(y)(x)))\cdot ((\left(\frac(y)(x) \right))^(\prime ))_(y)=\frac( 1)(2\sqrt(\frac(y)(x)))\cdot \frac(1)(x)=\]

\[=\frac(1)(2)\cdot \sqrt(\frac(x)(y))\cdot \sqrt(\frac(1)(((x)^(2))))=\frac (1)(2)\sqrt(\frac(x)(y\cdot ((x)^(2))))=\frac(1)(2\sqrt(xy))\]

Everything is shattered.

Manager No. 2

This butt is at once simpler and more folding, lower forward. More foldable, to that there is more action here, but simpler, to that there is no root here, moreover, the function is symmetrical to $x$ and $y$, tobto. As we remember $x$ and $y$ as missions, the formula does not seem to change. Tse respect had to be forgiven for the payment of private expenses, tobto. It's enough to damage one of them, and in the other one just remember $x$ and $y$ with the brushes.

Let's get to the point:

\[(((z)")_(x))=((\left(\frac(xy))(((x)^(2))+((y)^(2))+1) \ right ))^(\prime ))_(x)=\frac(((\left(xy \right))^(\prime ))_(x)\left(((x)^(2))+ ( (y)^(2))+1 \right)-xy((\left(((x)^(2))+((y)^(2))+1 \right))^(\prime ) )_(x))(((\left(((x)^(2))+((y)^(2))+1 \right))^(2)))\]

Let's get excited:

\[((\left(xy \right))^(\prime ))_(x)=y\cdot ((\left(x \right))^(\prime ))=y\cdot 1=y\ ]

Prote richly learn such a record of ignorance, we will write down the axis like this:

\[((\left(xy \right))^(\prime ))_(x)=((\left(x \right))^(\prime ))_(x)\cdot y+x\cdot ((\left(y \right))^(\prime ))_(x)=1\cdot y+x\cdot 0=y\]

In this rank, we once again switch over to the universality of the algorithm of private relatives: they didn’t care about them, if all the rules are set up correctly, you will be the one yourself.

Now let's take a look at one more private trick of our great formula:

\[((\left(((x)^(2))+((y)^(2))+1 \right))^(\prime ))_(x)=((\left((( x)^(2)) \right))^(\prime ))_(x)+((\left(((y)^(2)) \right))^(\prime ))_(x) +(((1)")_(x))=2x+0+0\]

Let's assume that we take away the dependence on our formula and take it away:

\[\frac(((\left(xy \right))^(\prime ))_(x)\left(((x)^(2))+((y)^(2))+1 \ right)-xy((\left(((x)^(2))+((y)^(2))+1 \right))^(\prime ))_(x))(((\left (((x)^(2))+((y)^(2))+1 \right))^(2)))=\]

\[=\frac(y\cdot \left(((x)^(2))+((y)^(2))+1 \right)-xy\cdot 2x)(((\left((( ( x)^(2))+((y)^(2))+1 \right))^(2)))=\]

\[=\frac(y\left(((x)^(2))+((y)^(2))+1-2((x)^(2)) \right))(((\ left(((x)^(2))+((y)^(2))+1 \right))^(2)))=\frac(y\left(((y)^(2)) -((x)^(2))+1 \right))(((\left(((x)^(2))+((y)^(2))+1 \right))^(2 )))\]

$x$ is reinstated. And in order to fix $y$ in the same viraz, let's not vikonuvat all the same sequence of diy, but rather with the symmetry of our vivid viraz - we just replace in our vivid viraz all $y$ by $x$ and navpak:

\[(((z)")_(y))=\frac(x\left(((x)^(2))-((y)^(2))+1 \right))((( ( \left(((x)^(2))+((y)^(2))+1 \right))^(2)))\]

For the rahunok of symmetry, they praised the whole viraz richly shvidshe.

nuance cherry

For the private ones, all standard formulas are used, which is the best for the private ones, but the same is true for the private one. With this, however, they blame their own specific features: if we respect $x$ privately, then if we take її for $x$, then we consider it as a constant, and to that її is similar to more expensive “zero”.

Like and at the same time with the most significant pokhіdnymi, private (one and the same) you can spoil a kіlkom in different ways. For example, the same construction, which was so well applauded, can be rewritten like this:

\[((\left(\frac(y)(x) \right))^(\prime ))_(x)=y\cdot ((\left(\frac(1)(x) \right)) ^(\prime ))_(x)=-y\frac(1)(((x)^(2)))\]

\[((\left(xy \right))^(\prime ))_(x)=y\cdot (((x)")_(x))=y\cdot 1=y\]

At once about those, from the other side, you can beat the formula in the form of a casual sum. As we know, there are more expensive sums of the dead. For example, let's write this:

\[((\left(((x)^(2))+((y)^(2))+1 \right))^(\prime ))_(x)=2x+0+0=2x \]

Now, knowing everything, let's try to improve with more serious usages, the shards of right private tricks are not surrounded by only rich terms and roots: trigonometry, logarithms, and display functions are used there. Now let's get busy.

Task with trigonometric functions and logarithms

Manager No. 1

We write the following standard formulas:

\[((\left(\sqrt(x) \right))^(\prime ))_(x)=\frac(1)(2\sqrt(x))\]

\[((\left(\cos x \right))^(\prime ))_(x)=-\sin x\]

Having mastered this knowledge, let's try to verse:

\[(((z)")_(x))=((\left(\sqrt(x)\cdot \cos \frac(x)(y) \right))^(\prime ))_(x )=((\left(\sqrt(x) \right))^(\prime ))_(x)\cdot \cos \frac(x)(y)+\sqrt(x)\cdot ((\left (\cos \frac(x)(y) \right))^(\prime ))_(x)=\]

Okremo write one change:

\[((\left(\cos \frac(x)(y) \right))^(\prime ))_(x)=-\sin \frac(x)(y)\cdot ((\left( \frac(x)(y) \right))^(\prime ))_(x)=-\frac(1)(y)\cdot \sin \frac(x)(y)\]

Turn to our design:

\[=\frac(1)(2\sqrt(x))\cdot \cos \frac(x)(y)+\sqrt(x)\cdot \left(-\frac(1)(y)\cdot \sin \frac(x)(y) \right)=\frac(1)(2\sqrt(x))\cdot \cos \frac(x)(y)-\frac(\sqrt(x))( y)\cdot \sin \frac(x)(y)\]

We all know about $x$, now let's get down to calculating $y$:

\[(((z)")_(y))=((\left(\sqrt(x)\cdot \cos \frac(x)(y) \right))^(\prime ))_(y )=((\left(\sqrt(x) \right))^(\prime ))_(y)\cdot \cos \frac(x)(y)+\sqrt(x)\cdot ((\left (\cos \frac(x)(y) \right))^(\prime ))_(y)=\]

Well, I know, I’m afraid one viraz:

\[((\left(\cos \frac(x)(y) \right))^(\prime ))_(y)=-\sin \frac(x)(y)\cdot ((\left( \frac(x)(y) \right))^(\prime ))_(y)=-\sin \frac(x)(y)\cdot x\cdot \left(-\frac(1)(( (y)^(2))) \right)\]

Let's turn to the end of the day and continue to see:

\[=0\cdot \cos \frac(x)(y)+\sqrt(x)\cdot \frac(x)(((y)^(2)))\sin \frac(x)(y) =\frac(x\sqrt(x))(((y)^(2)))\cdot \sin \frac(x)(y)\]

Everything is shattered.

Manager No. 2

Let's write down the formula we need:

\[((\left(\ln x \right))^(\prime ))_(x)=\frac(1)(x)\]

Now I'm sorry for $x$:

\[(((z)")_(x))=((\left(\ln \left(x+\ln y \right) \right))^(\prime ))_(x)=\frac( 1)(x+\ln y).((\left(x+\ln y \right))^(\prime ))_(x)=\]

\[=\frac(1)(x+\ln y)\cdot \left(1+0 \right)=\frac(1)(x+\ln y)\]

Found for $x$. Important for $y$:

\[(((z)")_(y))=((\left(\ln \left(x+\ln y \right) \right))^(\prime ))_(y)=\frac( 1)(x+\ln y).((\left(x+\ln y \right))^(\prime ))_(y)=\]

\[=\frac(1)(x+\ln y)\left(0+\frac(1)(y) \right)=\frac(1)(y\left(x+\ln y \right))\ ]

The task is over.

nuance cherry

Later, since the functions were not taken privately, the rules are overwritten by the same ones, regardless of whether they work with trigonometry, with roots or with logarithms.

The classic rules of work are always replaced by the standard ones, and at the same time, the sum of the retail, private and collapsible functions.

The rest of the formula is most often explained at the end of the day when the meeting is over with private holidays. Mi zustrіchaєmosya with them practically skrіz. There hasn’t yet been a city manager, so that we don’t get out there. But if we didn’t squirm with the formula, we still get one more benefit, and for ourselves, the peculiarity of the work with private walks. So we fix one change, the lines are constants. Zocrema, as we respect the privately lost virase $\cos \frac(x)(y)$ $y$, then $y$ itself is changed, and $x$ is overwritten with a constant. The same practice and navpaki. Її can be blamed for the bad sign, but bad as the constant itself is more like “zero”.

Everything should be brought to the point that private looks of one and the same viraz, but from different changes they can look differently. For example, marveling at such a virazi:

\[((\left(x+\ln y \right))^(\prime ))_(x)=1+0=1\]

\[((\left(x+\ln y \right))^(\prime ))_(y)=0+\frac(1)(y)=\frac(1)(y)\]

Task with demonstrative functions and logarithms

Manager No. 1

Let's write down the following formula:

\[((\left(((e)^(x)) \right))^(\prime ))_(x)=((e)^(x))\]

Knowing this fact, as well as the foldable functions, we can try to frighten. I believe in two different ways at once. The first and most obvious is the cost of the work:

\[(((z)")_(x))=((\left(((e)^(x))\cdot ((e)^(\frac(x)(y))) \right) )^(\prime ))_(x)=((\left(((e)^(x)) \right))^(\prime ))_(x)\cdot ((e)^(\frac (x)(y)))+((e)^(x))\cdot ((\left(((e)^(\frac(x)(y))) \right))^(\prime ) )_(x)=\]

\[=((e)^(x))\cdot ((e)^(\frac(x)(y)))+((e)^(x))\cdot ((e)^(\frac ) (x)(y)))\cdot ((\left(\frac(x)(y) \right))^(\prime ))_(x)=\]

Let's see this viraz:

\[((\left(\frac(x)(y) \right))^(\prime ))_(x)=\frac(((((x)"))_(x))\cdot yx .(((((y)"))_(x)))(((y)^(2)))=\frac(1\cdot yx\cdot 0)((((y)^(2) )) =\frac(y)((((y)^(2)))=\frac(1)(y)\]

Let's turn to our design and continue to see it:

\[=((e)^(x))\cdot ((e)^(\frac(x)(y)))+((e)^(x))\cdot ((e)^(\frac ) (x)(y)))\cdot \frac(1)(y)=((e)^(x))\cdot ((e)^(\frac(x)(y)))\left( 1 +\frac(1)(y)\right)\]

Everything, $x$ is covered.

However, as I said, at the same time we will try to protect my privacy in a different way. For whom respectfully so:

\[((e)^(x))\cdot ((e)^(\frac(x)(y)))=((e)^(x+\frac(x)(y)))\]

We write it down like this:

\[((\left(((e)^(x))\cdot ((e)^(\frac(x)(y))) \right))^(\prime ))_(x)=( (\left(((e)^(x+\frac(x)(y))) \right))^(\prime ))_(x)=((e)^(x+\frac(x)(y ) )))\cdot ((\left(x+\frac(x)(y) \right))^(\prime ))_(x)=((e)^(x+\frac(x)(y) ) )\cdot \left(1+\frac(1)(y) \right)\]

As a result, we took away the same amount of money, and the prote was charged as the smaller one. For whom to finish the bulk remember that when you finish the show, you can add up.

Now I'm sorry for $y$:

\[(((z)")_(y))=((\left(((e)^(x))\cdot ((e)^(\frac(x)(y))) \right) )^(\prime ))_(y)=((\left(((e)^(x)) \right))^(\prime ))_(y)\cdot ((e)^(\frac (x)(y)))+((e)^(x))\cdot ((\left(((e)^(\frac(x)(y))) \right))^(\prime ) )_(y)=\]

\[=0\cdot ((e)^(\frac(x)(y)))+((e)^(x))\cdot ((e)^(\frac(x)(y))) \cdot ((\left(\frac(x)(y) \right))^(\prime ))_(y)=\]

Let's sing one viraz okremo:

\[((\left(\frac(x)(y) \right))^(\prime ))_(y)=\frac(((((x)"))_(y))\cdot yx \cdot ((((y)"))_(y)))(((y)^(2)))=\frac(0-x\cdot 1)(((y)^(2))) =-\frac(1)((((y)^(2)))=-\frac(x)(((y)^(2)))\]

We sell the version of our external design:

\[=((e)^(x))\cdot ((e)^(\frac(x)(y)))\cdot \left(-\frac(x)(((y)^(2) )) \right)=-\frac(x)(((y)^(2)))\cdot ((e)^(x))\cdot ((e)^(\frac(x)(y) ))\]

It dawned on me that I could have lost my way in another way, I would have looked like this myself.

Manager No. 2

Fuck for $x$:

\[(((z)")_(x))=((\left(x \right))_(x))\cdot \ln \left(((x)^(2))+y \right )+x\cdot ((\left(\ln \left(((x)^(2))+y \right) \right))^(\prime ))_(x)=\]

Let's stop one viraz okremo:

\[((\left(\ln \left(((x)^(2))+y \right) \right))^(\prime ))_(x)=\frac(1)(((( x )^(2))+y)\cdot ((\left(((x)^(2))+y \right))^(\prime ))_(x)=\frac(2x)(( ((x)^(2))+y)\]

Sold solution of exterior design: $$

The axis is so clear.

Lost for analogy to know by $y$:

\[(((z)")_(y))=((\left(x \right))^(\prime ))_(y).\ln \left(((x)^(2)) +y \right)+x\cdot ((\left(\ln \left(((x)^(2))+y \right) \right))^(\prime ))_(y)=\]

One viraz, it’s ok, like a zavzhdi okremo:

\[((\left(((x)^(2))+y \right))^(\prime ))_(y)=((\left(((x)^(2)) \right) )^(\prime ))_(y)+(((y)")_(y))=0+1=1\]

Prodovzhuєmo virіshennya main designії:

Everything is covered. Like a bachite, fallow, depending on how the change is taken for differentiation, they appear absolutely different.

nuance cherry

The axis is a great example of how one and the same functions can be used in two different ways. Axis to wonder:

\[(((z)")_(x))=\left(((e)^(x))\cdot ((e)^(\frac(x)(y))) \right)=( (\left(((e)^(x)) \right))^(\prime ))_(x)\cdot ((e)^(\frac(x)(y)))+((e) ^(x))\cdot ((\left(((e)^(\frac(x)(y))) \right))^(\prime ))_(x)=\]

\[=((e)^(x))\cdot ((e)^(\frac(x)(y)))+((e)^(x))\cdot ((e)^(\frac ) (x)(y)))\cdot \frac(1)(y)=((e)^(x))\cdot ((e)^(^(\frac(x)(y)))) )\ left(1+\frac(1)(y) \right)\]

\[(((z)")_(x))=((\left(((e)^(x)).((e)^(\frac(x)(y))) \right)) ^(\prime ))_(x)=((\left(((e)^(x+\frac(x)(y))) \right))^(\prime ))_(x)=(( e)^(x+\frac(x)(y))).((\left(x+\frac(x)(y) \right))^(\prime ))_(x)=\]

\[=((e)^(x))\cdot ((e)^(^(\frac(x)(y))))\left(1+\frac(1)(y) \right)\ ]

When choosing different paths, the calculation could be different, but if it’s true, everything was done correctly, we’ll see it the same way. Prices are worthy of the classical, and private of the later ones. I’ll guess again from whom: it’s fallow, it’s like, what a change, I’ll take a good one, that’s it. differentiation, vіdpovіd can vyyti zovsіm raznoyu. Marvel:

\[((\left(\ln \left(((x)^(2))+y \right) \right))^(\prime ))_(x)=\frac(1)(((( x )^(2))+y)\cdot ((\left(((x)^(2))+y \right))^(\prime ))_(x)=\frac(1)(( (( x)^(2))+y)\cdot 2x\]

\[((\left(\ln \left(((x)^(2))+y \right) \right))^(\prime ))_(y)=\frac(1)(((( x )^(2))+y)\cdot ((\left(((x)^(2))+y \right))^(\prime ))_(y)=\frac(1)(( ((x)^(2))+y)\cdot 1\]

Nasamkinets for fixing all the material, let's try to fix two butts.

Task with a trigonometric function and a function with three changes

Manager No. 1

Let's write these formulas:

\[((\left(((a)^(x)) \right))^(\prime ))=((a)^(x))\cdot \ln a\]

\[((\left(((e)^(x)) \right))^(\prime ))=((e)^(x))\]

Let's now virishuvate our viraz:

\[(((z)")_(x))=((\left(((3)^(x\sin y)) \right))^(\prime ))_(x)=((3 )^(x.\sin y))\cdot \ln 3\cdot ((\left(x\cdot \sin y \right))^(\prime ))_(x)=\]

Okremo porahuemo such a design:

\[((\left(x\cdot \sin y \right))^(\prime ))_(x)=(((x)")_(x))\cdot \sin y+x((\ left(\sin y \right))^(\prime ))_(x)=1\cdot \sin y+x\cdot 0=\sin y\]

Prodovzhuєmo virishuvati vihіdny viraz:

\[=((3)^(x\sin y))\cdot \ln 3\cdot \sin y\]

This is the residual amount of private change $x$. Now I'm sorry for $y$:

\[(((z)")_(y))=((\left(((3)^(x\sin y)) \right))^(\prime ))_(y)=((3 )^(x\sin y))\cdot \ln 3\cdot ((\left(x\sin y \right))^(\prime ))_(y)=\]

Virishimo one viraz okremo:

\[((\left(x\cdot \sin y \right))^(\prime ))_(y)=(((x)")_(y))\cdot \sin y+x((\ left(\sin y \right))^(\prime ))_(y)=0\cdot \sin y+x\cdot \cos y=x\cdot \cos y\]

Virishuemo to the end of our design:

\[=((3)^(x\cdot \sin y))\cdot \ln 3\cdot x\cos y\]

Manager No. 2

At first glance, this butt can be folded, because there are three changes. Indeed, it is one of the simplest tasks for today's video tour.

Known by $x$:

\[(((t)")_(x))=((\left(x((e)^(y))+y((e)^(z)) \right))^(\prime ) )_(x)=((\left(x\cdot ((e)^(y)) \right))^(\prime ))_(x)+((\left(y\cdot ((e) ) ^(z)) \right))^(\prime ))_(x)=\]

\[=((\left(x \right))^(\prime ))_(x)\cdot ((e)^(y))+x\cdot ((\left(((e)^(y ) )) \right))^(\prime ))_(x)=1\cdot ((e)^(y))+x\cdot o=((e)^(y))\]

Now let's look at $y$:

\[(((t)")_(y))=((\left(x\cdot ((e)^(y))+y\cdot ((e)^(z)) \right))^ (\prime ))_(y)=((\left(x\cdot ((e)^(y)) \right))^(\prime ))_(y)+((\left(y\cdot ) ((e)^(z)) \right))^(\prime ))_(y)=\]

\[=x\cdot ((\left(((e)^(y)) \right))^(\prime ))_(y)+((e)^(z))\cdot ((\left (y \right))^(\prime ))_(y)=x\cdot ((e)^(y))+((e)^(z))\]

We knew the truth.

Now it's too much to know $z$:

\[(((t)")_(z))=((\left(x\cdot ((e)^(y))+((y)^(z)) \right))^(\prime ))_(z)=((\left(x\cdot ((e)^(y)) \right))^(\prime ))_(z)+((\left(y\cdot ((e )^(z)) \right))^(\prime ))_(z)=0+y\cdot ((\left(((e)^(z)) \right))^(\prime )) _(z)=y\cdot ((e)^(z))\]

We praised the third pokhidna, on which the vision of another task is completed again.

nuance cherry

Like a bachite, there is nothing folding in these two butts. The only thing, why we messed up, it’s because the foldable functions are often stagnant and stale, since we’re privately shy, we’ll have to change depending on the situation.

In the rest of the task, we were asked to work out the functions of three different ones. There is nothing terrible in tsomu, prote naprikintsі mi have crossed paths, that stench is one kind of one and it’s completely irritating.

Key moments

The rest of the vysnovki from today's video lesson is as follows:

1. Private expenses are taken into account as such, as if they were important, in order to take into account private expenses by one change, deciding all the changes that are included in this function, we take them as constants.
2. Pratsyyuyuchi s private pokhіdnymi vikoristovuєmo tі sami standard formulas, yak і z znichnym pokhіdnymi: suma, raznitsyu, pokhіdnu create і private і, zrozumіlo, pokhіdnu foldable functions.

Obviously, looking at one video lesson is not enough, so that I can expand on this topic again, so at once on my site, before this video, there is a set of tasks dedicated to this day’s topic - come in, zavantazhite, tweet this task and get in touch After all, you won’t have any everyday problems from private ones like sleeping or working independently. Obviously, this is far from the last lesson in modern mathematics, so go to our website, add VKontakte, subscribe to YouTube, put likes and follow us!

Private holidays stay at the heads of the functions of a small number of people. The rules of significance are exactly the same as for functions of one variable, with the only difference being that one of the variable traces is taken into account at the moment of differentiation by a constant (constant number).

Formula

Private dates for the function of two variables $ z (x, y) $ are written in the next look $ z "_x, z"_ y $ and follow the formulas:

Private holidays first order

$$ z"_x = \frac(\partial z)(\partial x) $$

$$ z"_y = \frac(\partial z)(\partial y) $$

Private trips in a different order

$$ z""_(xx) = \frac(\partial^2 z)(\partial x \partial x) $$

$$ z""_(yy) = \frac(\partial^2 z)(\partial y \partial y) $$

Zmishana is good

$$ z""_(xy) = \frac(\partial^2 z)(\partial x \partial y) $$

$$ z""_(yx) = \frac(\partial^2 z)(\partial y \partial x) $$

Private storage folding function

a) Let $ z(t) = f(x(t), y(t)) $, then similar folding functions will follow the formula:

$$ \frac(dz)(dt) = \frac(\partial z)(\partial x) \cdot \frac(dx)(dt) + \frac(\partial z)(\partial y) \cdot \frac (dy)(dt) $$

b) Let $ z (u, v) = z(x(u, v), y(u, v)) $, then repeat the following private functions after the formula:

$$ \frac(\partial z)(\partial u) = \frac(\partial z)(\partial x) \cdot \frac(\partial x)(\partial u) + \frac(\partial z)( \partial y) \cdot \frac(\partial y)(\partial u) $$

$$ \frac(\partial z)(\partial v) = \frac(\partial z)(\partial x) \cdot \frac(\partial x)(\partial v) + \frac(\partial z)( \partial y) \cdot \frac(\partial y)(\partial v) $$

Private holidays implicitly defined functions

a) Let $ F(x,y(x)) = 0 $, then $$ \frac(dy)(dx) = -\frac(f"_x)(f"_y) $$

b) Let $ F (x, y, z) = 0 $, then $ $ z "_x = - \frac (F"_x) (F"_z); z "_y = - \ frac (F"_y) ( F"_z) $$

Apply solution

butt 1

Find first-order private values ​​$z(x,y) = x^2 - y^2 + 4xy + 10$

Solution

For the value of a private variable in $ x $, we will use $ y $ as a constant value (number): $$ z"_x = (x^2-y^2+4xy+10)"_x = 2x - 0 + 4y + 0 = 2x+4y $$ For the value of a private function relative to $ y $, $ y $ is significant by a constant: $$ z"_y = (x^2-y^2+4xy+10)"_y = -2y+4x $$ If you don’t dare to break your task, then force yoga before us. We need a more detailed solution. You can learn about the progress of the calculation and take away the information. Tse dopomozhe every hour take the hall from the vikladach!

Vidpovid

$$ z"_x = 2x+4y; z"_y = -2y+4x $$

butt 2

Find private similar functions in a different order $ z = e ^ (xy) $

Solution

At the same time, it is necessary to know the first step, and then knowing them, you can know the steps of a different order. Important $ y $ constant: $$ z"_x = (e^(xy))"_x = e^(xy) \cdot (xy)"_x = ye^(xy) $$ Let's put now $ x $ constant value: $$ z"_y = (e^(xy))"_y = e^(xy) \cdot (xy)"_y = xe^(xy) $$ Knowing the first pokhіdnі, similarly we know others. We install $ y $ permanently: $$ z""_(xx) = (z"_x)"_x = (ye^(xy))"_x = (y)"_x e^(xy) + y(e^(xy))"_x = 0 + ye^(xy)\cdot(xy)"_x = y^2e^(xy) $$ Set $ ​​x $ constant: $$ z""_(yy) = (z"_y)"_y = (xe^(xy))"_y = (x)"_y e^(xy) + x(e^(xy))"_y = 0 + x^2e^(xy) = x^2e^(xy) $$ Now I have lost the knowledge of the zmіshanu pokhіdnu. You can differentiate $ z"_x $ with respect to $ y $, or you can differentiate $ z"_y $ with respect to $ x $, because of the theorem $ z""_(xy) = z""_(yx) $ $$ z""_(xy) = (z"_x)"_y = (ye^(xy))"_y = (y)"_y e^(xy) + y (e^(xy))"_y = ye^(xy)\cdot(xy)"_y = yxe^(xy) $$

Vidpovid

$$ z"_x = ye^(xy); z"_y = xe^(xy); z""_(xy) = yxe^(xy) $$

butt 4

Let $ 3x ^ 3z - 2z ^ 2 + 3yz ^ 2-4x + z-5 = 0 $ put an implicit function $ F (x, y, z) = 0 $. Know private events of the first order.

Solution

We write the function in the format: $F(x,y,z) = 3x^3z - 2z^2 + 3yz^2-4x+z-5 = 0$ $$ z"_x (y,z - const) = (x^3 z - 2z^2 + 3yz^2-4x+z-5)"_x = 3 x^2 z - 4 $$ $$ z"_y (x,y - const) = (x^3 z - 2z^2 + 3yz^2-4x+z-5)"_y = 3z^2 $$

Vidpovid

$$ z"_x = 3x^2 z - 4; z"_y = 3z^2; $$

Appointment 1.11 Let the function of two changers be set z=z(x,y), (x,y)D . Dot, mottled M 0 (x 0 ;y 0 ) - internal point of the area D .

Yakscho in D є such a neighborhood UM 0 points M 0 , which for all points

then point M 0 is called the local maximum point. And the meaning z(M 0 ) - local maximum.

And as for all points

then point M 0 is called the point of the local minimum of the function z(x,y) . And the meaning z(M 0 ) - Local minimum.

The local maximum and local minimum are called local extrema of the function z(x,y) . On fig. 1.4 explains the geometric change of the local maximum: M 0 - point to the maximum, to what is on the surface z = z(x, y) clear point C 0 know more than any other point C (Which has the maximum locality).

Respectfully, there are dots on the surface (for example, IN ), if you know more C 0 , ale qi dots (for example, IN ) not є "judicial" with a dot C 0 .

Zocrema, point IN confirms the understanding of the global maximum:

Similarly, the global minimum is determined:

The knowledge of global maximums and minimums will be discussed in paragraph 1.10.

Theorem 1.3(Necessary mind the extreme).

Let the function be set z = z (x, y), (x, y) D . Dot, mottled M 0 (x 0 ;y 0 D - point of local extremum.

What do you have z" x і z" y , then

Geometric confirmation is "obviously". What's next C 0 on (Fig. 1.4) to draw a dotically flat area, there "naturally" pass horizontally, i.e. under the hood 0° to axis Oh i to axis OU .

The same goes for a geometric change of private relatives (Fig. 1.3):

what it was necessary to bring.

Appointment 1.12.

What's next M 0 think (1.41), then it is called the stationary point of the function z (x, y) .

Theorem 1.4(Sufficient mind the extreme).

Let me ask z = z (x, y), (x, y) D , as there may be private events of a different order in the actual vicinity of the point M 0 (x 0 ,y 0 )D . And why M 0 - Stationary point Let's calculate:

The proof of the Vicorist theorem by those (Taylor's formula of the function of a number of variables and the theory of quadratic forms), which is not considered by any helper.

butt 1.13.

Go to the extremum:

Solution

1. We know the stationary points that break the system (1.41):

so we have found some stationary points. 2.

after Theorem 1.4, points have a minimum. And why

according to Theorem 1.4 at the point

Maximum. And why