# Geometric interpretation of frechet derivative in one variable

Given a function $f:(a,b) \to E$ where $E$ is a Banach space, then clearly when $f$ is differentiable, the derivative $f^\prime(x)$ can be seen as the tangent vector to the curve in $E$ graphed by $f$, at the point $f(x)$.

But the book I use says that value of the Frechet derivative of $f$ acting on $1$ i.e. $f^\prime(x)(1)$ or $Df(x)(1)$ is also the same as above definition.

I am confused. The infinitesimal "h" here is $1$. But isnt "h" supposed to be as small as possible and moreover it maybe that $|b-a| < 1$ ? How is this interpretation brought about?

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$h$ is not small, it's just some direction. The thing getting small is the parameter $s$ in $\lim_{s \rightarrow 0}\frac{1}{s}[f(x+sh)-f(x)]$. – Nick Alger Sep 3 '12 at 6:54
in that case...still $s$ shrinking to $0$ is same as saying $sh =t$ getting small right??The $"h"$ in the question is actually the $"t"$ here. So in general $"t"$ is supposed to be a vector in the neighbourhood of the zero vector so that $x+t$ is in the open domain right?domain here is $(a,b)$...so I was asking how $1$ crept into the scheme of things?? – Vishesh Sep 3 '12 at 7:11
How much would the function change if you walked 1 unit in the direction of $h$, assuming the function was replaced with it's best linear approximation at point $x$. – Nick Alger Sep 3 '12 at 7:52

When you start doing multivariable calculus you learn that the derivative of a function $f$ at a point $p$ should not be viewed as a number (or a vector) but as a linear map $Df(p):\ T_p\to T_{f(p)}$. Then of course one has to answer the question how this fits with the well-known derivative from one-variable calculus. The latter is defined by $$f'(p)=\lim_{h\to 0}{f(p+h)-f(p)\over h}$$ and is a number or a vector depending on the type of $f$. The equation defining $f'(p)$ may also be written in the denominator-free form $$f(p+h)-f(p)= h f'(p)+o(|h|)\qquad(h\to 0)\ ,$$ from which we immediately deduce that the Fréchet derivative of $f$ at $p$ is the linear map $df(p):\ h\mapsto h f'(p)$.
Now to your question: When the Fréchet derivative $df(p)$ (a linear map) is given to us, how can we extract from it the "old" derivative $f'(p)$? The answer is: Apply $df(p)$ to the "special" vector $h:=1\in{\mathbb R}^1$, and you get $df(p)(1)=1 f'(p)=f'(p)$.
Oh i get it now, thanks.I was trying to make sense of $h = 1$ in $f(p+h)$.I never looked at it as a mapping. – Vishesh Sep 3 '12 at 8:36