proof for derivative

Question

(f₁+f₂+f₃+...)^'=f1'+f2'+f3'+...

Answer 1

Proof:

 

(f₁ + f₂ + ... + f_n)' (x)

 

= lim_h→0 [(f₁ + f₂ + ... + f_n)(x+h) - (f₁ + f₂ + ... + f_n)(x)] / h

 

= lim_h→0 [f₁(x+h) + f₂(x+h)+ ... + f_n(x+h) - f₁(x) - f₂(x) - ... - f_n(x)] / h

 

= lim_h→0 [f₁(x+h) - f₁(x)]/h + lim_h→0 [f₂(x+h) - f₂(x)]/h + ... + lim_h→0 [f_n(x+h) - f_n(x)]/h

 

= f₁'(x) + f₂'(x) + ... + f_n'(x)

Answer 2

Fundamental Derivative Rule:

 

The derivative of the sum is the same as the sum of each term's derivatives.

 

d/dx (f₁ + f₂ + f₃) = d/dx (f₁) + d/dx (f₂) + d/dx (f₃)

 

 

d/dx is the notation for derivative, and this notation is a common factor of the terms.

 

 

For example:

 

(x³ + x⁴ + 7x⁸)' = (x³)' + (x⁴)' + (7x⁸)

 

3x² + 4x³ + 56x⁷ = 3x² + 4x² + 56x⁷