In finding the derivative of a piecewise function, why should the piecewise point be defined by the derivative? And how do you know whether it is differentiable or not before finding the derivative? And do you study math? It's too strong

In finding the derivative of a piecewise function, why should the piecewise point be defined by the derivative? And how do you know whether it is differentiable or not before finding the derivative? And do you study math? It's too strong

It's OK to use the derivative definition to determine the subsection point (you can also use the definition if it's not a subsection point, but you don't have to use the derivation formula. The key is whether the derivative function is continuous at the subsection point. If we use the derivation formula to calculate the derivative function on the right side of the subsection point, then the value of the subsection point x0 of future generations is taken as f '(x0), which is really

Why do we sometimes find the second derivative D for the second derivative of higher numbers ² y/dx ² = d/dx()′ × dy/dx ； And sometimes you do ask directly Sometimes the two are equal, sometimes they can't, For example, find the second derivative s = wcoswt. (s') '= - W ² sinwt; And apply the formula D / dt (s) ' × DS / DT is equal to - W ³ sinwt × Coswt. This is for god horse~ After reading your solution, I suddenly realized that I would write such a second-order formula. It may be affected by the following. In the derivation formula of inverse function, DX / dy = 1 / y '→ D ² x/dy ²= D (1 / y ') / dy = (why replace the denominator dy with DX here?) d (1 / y') / DX × dx/dy=-y''/(y') ³ ， Thus, the third power appears; There is no third independent variable here. Why not directly use d / Dy (DX / dy) = (1 / y ')'?

s(t)=cos wts'(t)=-wsin wts''(t)=[s'(t)]'=-w^2 cos wtd ² y/dx ² = d/dx()′ × Dy / DX is this the god horse formula ya? It should be: D ² y/dx ² = Dy '(x) / DX and y' (x) = D, y (x) / DX

The order of commutative successive integrals ∫ [0,1] DX ∫ [0,1-x] f (x, y) dy Explain the reason in the process

This is the area bounded by the line x + y = 1 and the two coordinate axes
Moreover, the integral field is symmetric about y = x, so it's OK to swap X and y
∫(0→1) dx ∫(0→1 - x) f(x,y) dy
= ∫(0→1) dy ∫(0→1 - y) f(x,y) dx

The order ∫ dy ∫ f (x, y) DX of commutative successive integrals of higher numbers, the first upper and lower limits are 1,0, and the second is 1-y, 0

After the exchange,
∫ dx∫ f(x,y)dy
First upper limit 1, lower limit 0
Second upper limit 1-x, lower limit 0

On the problem of double integration in higher numbers, ∫ (upper limit 1, lower limit 0) dy ∫ (upper limit 1, lower limit y) f (x, y) DX exchanges the order of integration

The area is surrounded by y = 0, y = 1, x = y, x = 1. Draw a graph
After the exchange order is
∫ (upper limit 1, lower limit 0) DX ∫ (upper limit x, lower limit 0) f (x, y) dy

Transform integration order ∫ (0,1) dy ∫ (- y, 1 + y ^ 2) f (x, y) DX

Original formula = ∫ - 1,0) DX ∫ - x, 1) f (x, y) dy + ∫ (0,1) DX ∫ (0,1) f (x, y) dy + ∫ (1,2) DX ∫ (√ (x-1), 1) f (x, y) dy