I saw this equation in a book and It was used as part of calculating something else in here and here Can I get any help about this equation
the statement is this:
suppose that: $$f: R\times R \rightarrow R$$ $$k: R\rightarrow R$$ we assume function G exits such that: $$f(x,k(x)+z) = f(x,k(x)) + G(x,z)z$$ prove that G is calculated by this formula: $G(x,z) = \int_{0}^{1} D_x(f(x,k(x)+yz)dy$
My problem is that the writers of the links I gave above use this equation and there is no proof or reference for that I want to know how this can be proved
Consider the function $h(s)=F(x,k(x)+sz)$. By the chain rule, its derivative is $$ h'(s)=(D_yF)(x,k(x)+sz)\cdot z,~~ \text{ which means } ~~ h'(s)=\left.\frac{\partial F(x,y)}{\partial y}\right|_{y=k(x)+sz}. $$ Now integrate this equation over $[0,1]$ and use that $z$ is constant to get $$ F(x,k(x)+z)=h(1)=h(0)+\int_0^1h'(s)\,ds=F(x,k(x))+\left[\int_0^1 (D_yF)(x,k(x)+sz)\,ds\right]\cdot z. $$ The notation in your claim seems slightly skewed in this light.