High Number Full Differentiation? 50, high number full differentiation?

Here's how, please refer to:

If it helps,

by the full differential formula.

Answering process.

The basic problems of calculus, the detailed process is as follows:

z=ln(2+x^2+y^2)

First, find the full differential into:

dz=(2xdx+2ydy) (2+x 2+y 2) then when x=2, y=1, there is:

dz(2,1)=(4dx+2dy)/(2+4+1)(4/7)dx+(2/7)dy.

Denote the partial differentiation with d: dz dx = 2x (2+x 2+y 2), dz dy = 2y (2+x 2+y 2).

Bring in x=2, y=1 to get it.

dz dx = 4 7, dz dy = 2 7 full differential dz = 4 7 dx +2 7 dy

Find the full differentiation of the function z=ln(2+x +y) at x=2,y=1;

Solution: dz=( z x)dx+( z y)dy=[2x (2+x +y )]dx+[2y (2+x +y )]dy;

Therefore, when x=2 and y=1, dz=(4 7)dx+(2 7)dy=(2 7)(2dx+dy);

The full differentiation of this function is.

dz = 2(xdx+ydy)/(2+x²+y²)。

You should know that the formula for total differentiation is dz=z'(x)dx+z'(y)dy, so these two derivatives, z, are found separately'(x)(x,y)=2x/(1+x^2+y^2), z'(y)(x,y)=2y (1+x 2+y 2), so z'(x)(1,2)=2/6=1/3,z'(y)(1,2)=4 6=2 3, so dz(1,2)=dx 3+2dy 3

First of all, find the full differentiation, dz 2x (1+x +y) dx+2y (1+x +y )dy brings x 1, y 2 in, dz 2 (1+1+4) dx+4 (1+1+4)dy 1 3dx+2 3dy

<>1. About the total differentiation of high numbers.

The result of Zheng Xuxing's quest is a=2, b=-1, and the answer is correct.

2.High number full differentiation, time finding, is mainly to use the differentiable definition. When it is differentiable, you can find the full differentiation.

3.In this question, we use the differentiable definition of the equivalent form, that is, the third line of the diagram shouts the differentiable definition of the form of good luck.

4.The third line in my diagram, which can be differentiated, corresponds to the sixth row in the diagram, and the part corresponding to the pencil is the two partial derivatives.

The specific high number full differentiation, the result is a=2, b=-1 The answer is correct, and the detailed steps and instructions for finding it are shown above.

First of all, the formula is deformed.

This form is very similar to the derivative, but it is a difference coefficient, multiplying the numerator and denominator by 2 and 3 respectively, and the next step is to find the derivative at x 0, because the derivative of the even function f(x) at x 0 is 0, so the final answer is 0

f is an even function, =>f'(0)=0

lim(t->0) [f(2t) -f(3t)] t (0 0 numerator denominator derivative respectively).

lim(t->0) [2f'(2t) -3f'(3t)]=-f'(0)=0

It is possible to calculate the points that follow!

Note:There should be no x in your integral expression numerator

Let x=2sinu, then Yuvolt dx=2cosudu, bring in the know, the original integral = 1 2) 1 dx

1/2)∫2cosudu/2sinucosu(1/2)∫du/sinu

1/2)∫cscudu

1/2)ln(cscu-cotu)+c

1 2) LN+C is certified!

Solution: Let Z=Arctanu V, and U=X+Y, V=1-Xy, so dz=[1 (1+(U V) 2) (1 V)]du+[1 (1+(U V) 2) (U V2)]dv

And because du=dx+dy, dv=-ydx-xdy is substituted for dz.

dz=[1/(1+(x+y/1-xy)^2)×(1/1-xy)](dx+dy)+[1/(1+(x+y/1-xy)^2)×(x+y/(1-xy)^2)](ydx-xdy)

Simplify. dz=1/(1+x^2)dx+1/(1+y)^2dy