h In what context was F G used ?
NothingA b c d e f g h does not have a meaning. They are the first 8 letters of the Engish alphabet.
a,a,g,g,h,h,g,f,f,d,d,s,s,a
Recall that a linear transformation T:U-->V is one such that 1) T(x+y)=T(x)+T(y) for any x,y in U 2) T(cx)=cT(x) for x in U and c in R All you need to do is show that differentiation has these two properties, where the domain is C^(infinity). We shall consider smooth functions from R to R for simplicity, but the argument is analogous for functions from R^n to R^m. Let D by the differential operator. D[(f+g)(x)] = [d/dx](f+g)(x) = lim(h-->0)[(f+g)(x+h)-(f+g)(x)]/h = lim(h-->0)[f(x+h)+g(x+g)-f(x)-g(x)]/h (since (f+g)(x) is taken to mean f(x)+g(x)) =lim(h-->0)[f(x+h)-f(x)]/h + lim(h-->0)[g(x+h) - g(x)]/h since the sum of limits is the limit of the sums =[d/dx]f(x) + [d/dx]g(x) = D[f(x)] + D[g(x)]. As for ths second criterion, D[(cf)(x)]=lim(h-->0)[(cf)(x+h)-(cf)(x)]/h =lim(h-->0)[c[f(x+h)]-c[f(x)]]/h since (cf)(x) is taken to mean c[f(x)] =c[lim(h-->0)[f(x+h)-f(x)]/h] = c[d/dx]f(x) = cD[f(x)]. since constants can be factored out of limits. Therefore the two criteria hold, and if you wished to prove this for the general case, you would simply apply the same procedure to the Jacobian matrices corresponding to Df.
f'(g(h(x)))*g'(h(x))*h'(x) where the prime denote a derivative with respect to x.
h In what context was F G used ?
3 = the Holy Trinity - Father, Son and Holy Ghost
F. G. H. Salusbury has written: 'George V and Edward VIII'
What 1s 14kp
Yes. Once u loosen up your face. Trololololololol Hui G G G G G D S X Y C H T C Jy F J F F H G Dusidhfitfujr
F F g h u y t h f s j
f = 54