B Derivation of exact differential

[kidsasd987]

Exact differential of a scalar function f takes the form of

∇f⋅dr=Σ∂_ifdx_i (where dr is a vector)
f:R->Rⁿand I am not sure why this equation is valid in the sense that if we integrate the equation,

∫∇f⋅dr=∫{Σ∂_ifdx_i}
∫df=∫{Σ∂_ifdx_i}

the above equation is true because integration is a linear operator, and if we think of R.H.S only,

∫{Σ∂₁fdx₁}+∫{Σ∂_i≠1fdx_i≠1}
=f+c(x₂,x₃...,x_n)+∫{Σ∂_i≠1fdx_i≠1}

and if we think of each integration of differential w.r.t variable x₁,x₂,x₃ and so on, it should generate f+c respectively as the result of each integration.

therefore R.H.S has to be

nf+C(x₁,x₂,x₃...,x_n) but according to textbook, it says we don't add up each integration but we compare them to eradicate constants and "merge" each equation to one right answer, f.

I am a little bit confused about how to interpret the integration within the quotation mark ∫df="∫"{Σ∂_ifdx_i} because it seems it is linearly applied to each of partial differential, but does not spit out nf(n number of partial differentials so there must be n number of integrations on them so adding them up would give nf+C(x₁,x₂,x₃...,x_n)

Please enlighten me. I would really appreciate

 

[Mgcini Keith Phuthi]

As you noted, the gradient operator takes a scalar in R and gives a vector in $$R^n$$ and each derivative of f in the sum gives a component of the resulting vector. The key here is that the differential operators all act on f and they act differently. For instance you can have f = xy, taking the 2D gradient gives df/dx = y and df/dy = x. The scalar function f that produces the derivative is not the sum of derivatives which is 2*xy but is instead f = xy. This is what the book means by compare, eradicate and merge. You know there must be an xy term from integrating the x derivative and similarly for the y derivative but the result is not their sum in this case, you can test this by differentiating it again to see if it gives the same gradient.