come Set in (a, b), f '(x) = g' (x), then the following formulas must be true? A.(∫f（x）dx)’=(∫g（x）dx)’ B.∫f’（x）dx=∫g’（x）dx I know a is right, I can't understand it, or B. that expression doesn't make any sense at all?

The derivatives of two functions are the same, and the two functions are not necessarily the same
∫f’（x）dx=f(x)+C
C is a constant that can be changed
(∫f（x）dx)’=(∫g（x）dx)’
This simplification should be
F (x) = g (x) seems wrong

Ask an after-school question in the book of advanced mathematics to find the direction derivative This is in the author's solution How can you tell that this corner is the corner of the third quadrant

What I understand is that a and B are positive, and because we are looking for the direction of the inner normal, the angle is in the third quadrant
Explain in detail, but you probably know
Dy / DX = - FX / FY, so the slope of the normal is FY / FX = (a ^ 2Y) / (b ^ 2x)
Directional derivative = - (2x / A ^ 2 * cosa + 2Y / b ^ 2 * Sina)
Cosa = B ^ 2x / root sign (a ^ 4Y ^ 2 + B ^ 4x ^ 2), bring in the values of X and y, and sina is the same
Because it's in the third quadrant, so the original formula=
(1 / radical 2) * radical (a ^ 2 + B ^ 2) / AB

In high numbers, partial derivatives exist. Can we deduce the existence of directional derivatives?

The existence of partial derivatives is a necessary condition for derivability, and the continuity of partial derivatives is a sufficient condition for derivability. Of course, this is for derivability
The partial derivative exists, and the directional derivative exists~

Relationship between partial derivative and directional derivative Is there any relationship between partial derivative and directional derivative in Higher Mathematics? If so, what is the relationship?

Of course it does. The partial derivative is the directional derivative along the coordinate axis
The partial derivative is the partial derivative of the coordinate axis, and the directional derivative can be in any direction

What is the derivative of the unit vector over time t

1. If it is a rectangular coordinate system, the unit vectors I, J and K, because they are constant vectors, and the derivative is equal to 0; 2. If it is the unit vector of force, strength,,, etc. at any point in the physical problem, because the orientation of this unit vector in space is not fixed, as long as the physical quantity of each point in space changes with time, the unit vector

Derivative of higher number Let f (x) = cosx, prove (cosx) '= - SiNx

f'(x)=f(x+△x)=cos(x+△x) △x->0
When △ X - > 0, Lim [cos (x + △ x) - cosx] / △ x = LIM (cosxcos △ x-sinxsin △ x-cosx) / △ x = Lim sinxsin △ X / △ x
When △ X - > 0, sin △ X / △ x = 1
So (cosx) '= - SiNx

come Set in (a, b), f '(x) = g' (x), then the following formulas must be true? A.(∫f（x）dx)’=(∫g（x）dx)’ B.∫f’（x）dx=∫g’（x）dx I know a is right, I can't understand it, or B. that expression doesn't make any sense at all?