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let f be a function map S to R and f is differentiable on an open set S. S belongs to R^2 and f has a local min or max at point a.a belongs to S.

to prove the gradient of f at a equals to 0

Your question is easier to answer if S and R are open intervals containing a.

But here is what you do to reduce the problem to (one dimensional) R^1.

Say the co-ordinates of a = (c,d). Consider the following two differentiable functions:

F(x) = f(x,d) (fix the y-co-ordinate of the point a)

G(y) = f(c,y) (fix the x-co-ordinate of the point a)

If we can prove that both F'(c)=0 and G'(d)=0 then we are done because the

grad(f) (a) = grad(f) (c,d) = (F'(c), G'(d))

So what I want to prove is:

Assuming c is a local max (and you have to do a different but similar proof for local min) of a differentiable function F from an open interval that contains c to an open set in R^1 (or the real line).

Key ideas:

1) Being a local max, we can assume that if the open interval is very small, F(c) >= F(x) for all x in that small tiny interval. Which implies that F(x) - F(c) is negative (actually non-positive is a more accurate word mathematically).

2) The left limit of the difference quotient: lim {x->c^-} (F(x)-F(c))/(x-c) is greater than or equal to zero because the numerator is negative and the denominator is negative (x are all taken to be on the left of c).

3) The right limit of the difference quotient: lim {x->c^+} (F(x)-F(c))/(x-c) is less than or equal to zero because the numerator is negative and the denominator is positive (x are all taken to be on the right of c).

But F'(x) which we know is the limit of the difference quotient exists (because F is differentiable). But together with (2) and (3), this limit of the difference quotient must be zero. ( lim {x->c^-} (F(x)-F(c))/(x-c) = lim {x->c} (F(x)-F(c))/(x-c) = lim {x->c^+} (F(x)-F(c))/(x-c) , i.e. left limit = limit = right limit)

Now put all these little arguments together and you got a proof that the gradient of f at a is equal to (0,0).

Your question is more than the typical calculus. It requires analysis and it's more of multivariable calculus which folks in America do not consider Calculus, unless you mention Calculus 3.

Calculus

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I can answer all calculus question. I am a Math Lecturer and I teach Math in a College, usually Calculus 1, Calculus 2 and Calculus 3 and Linear Algebra.

I have been teaching in this college for more than 12 years. Prior to that, I taught for three years as Visiting Assistant Professor.**Education/Credentials**

I got my doctorate from Univ. of Rochester in Algebraic Toppology. I got a MA from Univ of Rochester, an MA from York Univ. in Toronto and almost did my M.Sc in the National Univ. of Singapore.