# Directional Derivative ... Tapp, Definition 3.3 ...

#### Peter

##### Well-known member
MHB Site Helper
I am reading Kristopher Tapp's book: Differential Geometry of Curves and Surfaces ... and am currently focused on Chapter 3: Surfaces ... and in particular on Section 1: The Derivative of a Function from $$\displaystyle \mathbb{R}^m$$ to $$\displaystyle \mathbb{R}^n$$ ... ...

I need help in order to fully understand Definition 3.3 ... ...

In the above text from Tapp we read the following:

$$\displaystyle df_p(v) = \displaystyle \lim_{ t \to 0 } \frac{ f(p + tv) - f(p) }{t} = (f \circ \gamma) (0)$$ ... ... "

Can someone please demonstrate (formally and rigorously) exactly why/how

$$\displaystyle \displaystyle \lim_{ t \to 0 } \frac{ f(p + tv) - f(p) }{t} = (f \circ \gamma) (0)$$ ... ...?

Hope someone can help ...

Peter

Last edited:

#### Opalg

##### MHB Oldtimer
Staff member
Can someone please demonstrate (formally and rigorously) exactly why/how

$$\displaystyle \displaystyle \lim_{ t \to 0 } \frac{ f(p + tv) - f(p) }{t} = (f \circ \gamma)' (0)$$ ... ...?
By definition, $(f \circ \gamma) (t) = f(\gamma(t)) = f(p+vt)$. Also by definition, $$(f \circ \gamma)' (0) = \lim_{t\to0}\frac{(f \circ \gamma) (t) - (f \circ \gamma) (0)}t = \lim_{t\to0}\frac{f(p+vt) - f(p)}t.$$ I guess the point here is that $f\circ\gamma$ is a function of a real variable $t$. So its derivative is given by the definition of single-variable calculus.

#### Peter

##### Well-known member
MHB Site Helper
By definition, $(f \circ \gamma) (t) = f(\gamma(t)) = f(p+vt)$. Also by definition, $$(f \circ \gamma)' (0) = \lim_{t\to0}\frac{(f \circ \gamma) (t) - (f \circ \gamma) (0)}t = \lim_{t\to0}\frac{f(p+vt) - f(p)}t.$$ I guess the point here is that $f\circ\gamma$ is a function of a real variable $t$. So its derivative is given by the definition of single-variable calculus.

Thanks Opalg ...