I want to know whether there is any such theorem in maths

[Atharva]

Theorem:-
For any quadratic function f(x), the mean of the derivative of any two points is equal to the derivative of mean of those two points.


Let f(x) be a real valued quadratic function defined as:-
f(x)=ax^2 +bx +c
Then, f'(x)= 2ax+b
Let's consider a interval [i , j] that is defined under the domain of the function
Thus,
f'(i)=2ai+b . And
f'(j)=2aj+b
Then,
(f'(i)+f'(j))/2 = a(i+j)+b -(1)
Now, let x=(i+j)/2
f'(x)=f'((i+j)/2)=2a((i+j)/2)+b
= a(i+j)+b - (2)
From (1) & (2) we get
(f'(i)+f'(j) )/2 = f'((i+j)/2)

 

[fresh_42]

Theorem:-
For any quadratic function f(x), the mean of the derivative of any two points is equal to the derivative of mean of those two points.


Let f(x) be a real valued quadratic function defined as:-
f(x)=ax^2 +bx +c
Then, f'(x)= 2ax+b
Let's consider a interval [i , j] that is defined under the domain of the function
Thus,
f'(i)=2ai+b . And
f'(j)=2aj+b
Then,
(f'(i)+f'(j))/2 = a(i+j)+b -(1)
Now, let x=(i+j)/2
f'(x)=f'((i+j)/2)=2a((i+j)/2)+b
= a(i+j)+b - (2)
From (1) & (2) we get
(f'(i)+f'(j) )/2 = f'((i+j)/2)

Yes, it is called linearity.

Your question is, if $\frac{1}{2}\cdot (D_if+D_jf) = D_{\frac{1}{2}(i+j)}f$ holds for quadratic $f$. Since the derivative $F$ of a quadratic function is a linear function, and evaluated at a certain point, it's an affine linear function, we have to decide, whether $\frac{1}{2}(F(i)+F(j)) = F(\frac{i+j}{2})$. Now this is true for all linear or constant functions $F$, for sums of those, and thus for affine linear functions as $F(p)$.

Linearity is the answer to your question.