Y 'is dy / DX. And y' 'is d ^ 2Y / DX ^ 2. Then why are the [second derivative] obtained by these two methods sometimes different Example 1: it is known that DX / dy = 1 / y '. Why does d ^ 2x / dy ^ 2 = [D (DX / dy) / dy] * DX / dy = - y' '/ y' ^ 3 instead of directly equal to - y '' / y '^ 2? Why does [D (DX / dy) / dy] multiply by DX / dy Example 2: S = asinwt. DS / dt = aw2coswt. D ^ 2S / dt ^ 2 = - aw ^ 2sinwt. How can example 2 not be calculated like example 1 If you know the algorithm of use case 1 and use case 2, please tell us the judgment method,

Y 'is dy / DX. And y' 'is d ^ 2Y / DX ^ 2. Then why are the [second derivative] obtained by these two methods sometimes different Example 1: it is known that DX / dy = 1 / y '. Why does d ^ 2x / dy ^ 2 = [D (DX / dy) / dy] * DX / dy = - y' '/ y' ^ 3 instead of directly equal to - y '' / y '^ 2? Why does [D (DX / dy) / dy] multiply by DX / dy Example 2: S = asinwt. DS / dt = aw2coswt. D ^ 2S / dt ^ 2 = - aw ^ 2sinwt. How can example 2 not be calculated like example 1 If you know the algorithm of use case 1 and use case 2, please tell us the judgment method,

Example 1 [DX / dy = 1 / y '], example 2 [S = asinwt. DS / dt = aw2coswt]
We can know that 1 is y (x) is a function of X, and 2 is s (T) is a function of T. therefore, 1 must use the reciprocal method to find the guide for y, and 2 uses the normal method~
1 is y (x) is a function of X, so the reciprocal method must be used to find the guideline of 1 for y
d^2x/dy^2
=d(x')/dy
=d(dx/dy)/dy
=d(dx/dy)/dx*dx/dy

In the derivative, d ^ 2Y / DX ^ 2 is to find the second derivative. What does d ^ 2Y / DX, dy / DX ^ 2 mean

The latter two are meaningless

The simple problem of finding the second derivative of implicit function is known dy / DX = - X / y to find d ^ 2Y / DX ^ 2 The concept of implicit function seeking second derivative is a little confused. It is known that dy / DX = - X / y seeking d ^ 2Y / DX ^ 2 (second derivative) I think d ^ 2Y / DX ^ 2 = D (dy / DX) / DX = D (- X / y) / DX = - 1 / Y (take y as a constant) The correct answer is: D ^ 2Y / DX ^ 2 = - (y-xy ') / y ^ 2 = - 1 / y ^ 3 I don't know what's wrong (the original equation is x ^ 2 + y ^ 2-1 = 0, continuous at point (0,1))

When we find the derivative dy / DX of the implicit function f (x, y) = 0, we regard y as
　　dy/dx=-x/y
When you find the second derivative, you still look at y = y (x), so,
　d^2y/dx^2 = d(dy/dx)/dx = d(-x/y)/dx = -(y-xy')/y^2 = …….

In the second derivative of the parametric equation, d ^ 2Y / DX ^ 2 = (D / DX) (dy / DX) = (D / DT) (1 / DX / DT) (dy / DX), is a number? Or a symbol similar to addition, subtraction, multiplication and division? What about D / dt?

That's how it works
y'=dy/dx=(dy/dt)/(dx/dt)
y"=d(y')/dx=d(y')/dt/(dx/dt)
D is differential, Dy is differential to y, DX is differential to x, DT is differential to t
The derivative is regarded as the quotient of two differentials, that is, y '= dy / DX. If the numerator and denominator are divided by DT at the same time, it is y' = (dy / DT) / (DX / DT)
Do the same for y 'and you get the second derivative

It is proved that the images of functions y = ax and y = A-X (a > 0 and a ≠ 1) are symmetrical about the Y axis

Take any point a (m, am) on the image of function y = ax, then the coordinate of point a about symmetric point B of Y axis is (- m, am),
For the function y = A-X, let x = - m, you can get y = am, so point B is on the image of function y = a-x,
Therefore, the images of the functions y = ax and y = A-X (a > 0 and a ≠ 1) are symmetrical about the Y axis

The x power of function y = 2 and the image of function f (x) are symmetrical about the Y axis, then f (x) is equal to? The answer is to the - x power of 2 Process thank you

On Y-axis symmetry
f(-x)=y=2^x=2^[-(-x)]
f(x)=2^-x