Why do we take 3A as a whole to calculate the derivative of implicit function x ^ 3 + y ^ 3-3axy = 0? 3x^2+3y^2*y~-3ay-3axy~=0 Divide it all by three x^2+y^2*y~-ay-axy~=0 Move those with y to one side y^2*y~-axy~=ay-x^2 =(ay-x^2)/(y^2-ax)

Why do we take 3A as a whole to calculate the derivative of implicit function x ^ 3 + y ^ 3-3axy = 0? 3x^2+3y^2*y~-3ay-3axy~=0 Divide it all by three x^2+y^2*y~-ay-axy~=0 Move those with y to one side y^2*y~-axy~=ay-x^2 =(ay-x^2)/(y^2-ax)

Because in this equation, only x, y are unknowns and a is a constant

Given the derivative equation x / (1 + x ^ 2) ^ 1 / 2, find the original equation f (x)

A:
f'(x)=x/√(1+x^2)
Points:
f(x)=∫ x/√(1+x^2) dx
=(1/2) 1/√(1+x^2) d(1+x^2)
=(1/2)*2*√(1+x^2)+C
f(x)=√(1+x^2)+C

How to find the derivative when f '(x) exists in the expression of F (x)?
For example, how to find the derivative f (x) = x2 + 2F '(- 1 / 3) x

There are two cases: first, when the independent variable in F '(x) is not determined, the independent variable is variable;
2、 When the independent variable in F '(x) has not been determined, the independent variable has been assigned
In the first case, let's take an example to illustrate: F (x) = x2 + 2F '(x) X. in the derivation process, f' (x) is regarded as the function of X (in fact, that's the case), and it can be solved according to the derivation rule
Left = f '(x), right = 2x + 2F' (x) x + 2F '(x), the derivative results contain the first and second order derivatives, which is normal
So f '(x) = 2x + 2F' (x) x + 2F '(x)
In the second case, take an example to illustrate: F (x) = x2 + 2F '(- 1 / 3)) X. it has been said that f' (x) is a function. When the independent variable is assigned, such as f '(- 1 / 3)), then it is a constant. According to the derivation rule, we can get
Left = f '(x), right = 2x + 2F' (- 1 / 3)
Grasp the basic concept of function can solve all the problems of derivation