A problem about the derivative of univariate function The question is still unclear Taking y as an independent variable and X as a dependent variable, the transformation equation is proved {(dy/dx) * [(dy)^3/d(x^3)]} - 3 {[(dy)^2/d(x^2)] ^2} = x One question: The answer is: [(dy)^3/d(x^3)] = - d/dy{(dx/dy)^(-3)*[(dx)^2/d(y^2)]}*(dy/dx) ={3(dx/dy)^(-4)*[(dx)^2/d(y^2)]^2-[(dx/dy)^(-3)]*(dx)^3/d(y^3)}*（dx/dy）^(-1) =3(dx/dy)^(-5)*[(dx)^2/d(y^2)]^2-(dx/dy)^(-4)*(dx)^3/d(y^3) But how do I figure it out =3(dx/dy)^(-5)*[(dx)^2/d(y^2)]^2-(dx/dy)^(-4)*(dx)^3/d(y^3)*[(dx)^2/d(y^2)] In other words, it is multiplied by one more [(DX) ^ 2 / D (y ^ 2)] I do this: let DX / dy = t So (Dy) ^ 3 / D (x ^ 3) =-d/dy[t^(-3)*t']*t^(-1) =[3t^(-4)*t'-t^(-3)*t'']*t'*t^(-1) =[3t^(-5)*t'^2-t^(-4)*t''*t']*t' I've done it several times. Why is there more t 'in the last step Please kindly answer it. Don't worry. The steps are a little troublesome. Take your time Thanks a lot^^~~

A problem about the derivative of univariate function The question is still unclear Taking y as an independent variable and X as a dependent variable, the transformation equation is proved {(dy/dx) * [(dy)^3/d(x^3)]} - 3 {[(dy)^2/d(x^2)] ^2} = x One question: The answer is: [(dy)^3/d(x^3)] = - d/dy{(dx/dy)^(-3)*[(dx)^2/d(y^2)]}*(dy/dx) ={3(dx/dy)^(-4)*[(dx)^2/d(y^2)]^2-[(dx/dy)^(-3)]*(dx)^3/d(y^3)}*（dx/dy）^(-1) =3(dx/dy)^(-5)*[(dx)^2/d(y^2)]^2-(dx/dy)^(-4)*(dx)^3/d(y^3) But how do I figure it out =3(dx/dy)^(-5)*[(dx)^2/d(y^2)]^2-(dx/dy)^(-4)*(dx)^3/d(y^3)*[(dx)^2/d(y^2)] In other words, it is multiplied by one more [(DX) ^ 2 / D (y ^ 2)] I do this: let DX / dy = t So (Dy) ^ 3 / D (x ^ 3) =-d/dy[t^(-3)*t']*t^(-1) =[3t^(-4)*t'-t^(-3)*t'']*t'*t^(-1) =[3t^(-5)*t'^2-t^(-4)*t''*t']*t' I've done it several times. Why is there more t 'in the last step Please kindly answer it. Don't worry. The steps are a little troublesome. Take your time Thanks a lot^^~~

The essence of your approach is the same as the answer
However, you made a mistake in deriving D / dy [T ^ (- 3) t ']
You see, its argument here has always been y, not X
In other words, t is a function of Y, not a composite function of X, so it doesn't have to be multiplied by T '
That place that day
If (DX / dy) ^ (- 1) = u,
Here y is in the position of the argument, so u is a function of Y,
D / DX [(DX / dy) ^ (- 1)] = Du / DX - → the independent variable in this step is X
So the final independent variable of this formula is X - → is u to take the derivative of X
And because y is a function of X
So u is a composite function of X
So D ² y/dx ²= d/dx[(dx/dy)^(-1)]=du/dx=(du/dy)(dy/dx)

A problem of function and derivative It is proved that when a > ln 2-1 and x > 0, e ^ x > x ^ 2-2ax + 1

E ^ x > x ^ 2-2ax + 1, i.e. e ^ x-x ^ 2 + 2aX > 1 Let f (x) = e ^ x-x ^ 2 + 2aX find the derivative f '(x) = e ^ x-2x + 2a, and then find the derivative f' '(x) = e ^ X-2, so that e ^ x = 2 = = > x = LN2 ∈ (0, LN2), f' '(x) ln2-1 ∈ a-ln2 + 1 > 0, i.e. f' (x) > 0. F (x) is the increasing function. F (x) > F (0) = e ^ 0 = 1, i.e. E. e ^ x-x ^ 2 + 2aX > 1, i.e

Find an example of a univariate function using derivative to find the maximum value of limit!

y=x^2
Y '= 2x when x = 0, y = 0, when x < 0, y' < 0, when x > 0, y '> 0
Therefore, (0,0) minimum value point, minimum value 0, no maximum value

How to explain in VC + + that a function is the derivative of another function for(i=0;i

VC has no such mechanism and can only write two functions respectively. One represents the function and the other represents the derivative of the function
As for your code, array t stores sampling points to represent the value of independent variables; Array x stores the original function value at the sampling point, that is, the cosine function; The array Z stores the derivative value at the sampling point; And array t_ Half stores the midpoint of two adjacent sampling points

Factorization method for solving univariate cubic equation Equation: a^3-2a^2-a+7=5 The unknowns are a 3. Everyone

It is impossible for mathematicians to find a root formula for a cubic equation of one variable,
Therefore, the univariate cubic equation can only be solved by factorization
a^3-2a^2-a+7=5
a^3-2a^2-a+2=0
a^3-a-2a^2+2=0
a(a^2-1)-2(a^2-1)=0
(a-2)(a+1)(a-1)=0
So there are three solutions: a = 2, a = - 1 and a = 1

On the solution of univariate cubic equation (general solution) For the equation AX ^ 3 + BX ^ 2 + CX + D = 0, is there a root formula, and the root formula is expressed by constants a, B, C and D?

The root formula of the univariate cubic equation is called "Cardano formula". The general form of the univariate cubic equation is X3 + SX2 + TX + U = 0. If we make an abscissa translation y = x + S / 3, then we can eliminate the quadratic term of the equation. Therefore, we only need to consider the cubic equation in the form of X3 = PX + Q. suppose the solution X of the equation can