Computing differentials of a quotient (product) of differential functions 
Problem Statement: Let $f:U\rightarrow \mathbb{R}^{n}$ and $g:U\rightarrow \mathbb{R}^{p}$ be differentiable ($U\subset \mathbb{R}^{m}$ open). Define: $$h(\mathbf{x}):U\setminus \ker{f}\rightarrow \mathbb{R}^{p},$$ $$\ \ \ \ \mathbf{x}\mapsto \frac{g(\mathbf{x})}{\lVert f(\mathbf{x})\rVert},$$ with $\lVert\cdot\rVert$ the euclidean norm in $\mathbb{R}^{p}$. Compute $dh(\mathbf{x})[\mathbf{v}]$, for each $\mathbf{v}\in \mathbb{R}^{m}$

So I am trying to work this problem out, but am a little confused because I would think that I need to use the product/quotient rule to determine $dh$, but we have not learned that in my Analysis course yet. I first took $\phi(\mathbf{x})=\lVert f(\mathbf{x})\rVert=\sqrt{\langle f(\mathbf{x}),f(\mathbf{x})\rangle}$, where $\langle\  ,\ \rangle$ is the standard inner product in $\mathbb{R}^{m}$, then I computed $d\phi(\mathbf{x})[\mathbf{v}]=\frac{1}{\lVert f(\mathbf{x})\rVert}\langle df(\mathbf{x})[\mathbf{v}],f(\mathbf{x})\rangle$. I am not sure where to go from here. Should I simply apply regular the product rule with $g(\mathbf{x})$ and $\frac{1}{\lVert f(\mathbf{x})\rVert}$?
Also, when computing differentials in real analysis, must we be precise in doing so by using the $\epsilon$-$\delta$ definition of derivative? In computing $d\phi(\mathbf{x})[\mathbf{v}]$, I just applied the standard limit definition to find the directional derivative in the direction of $\mathbf{v}\in\mathbb{R}^{m}$.
Any hints to lead me in the right direction are appreciated!
 A: I don't really understand your comment in the second to the last paragraph. Anyway, here's my computation using the definition
\begin{align}
\lim_{t\rightarrow 0} \frac{h(\mathbf{x}+t\mathbf{v})-h(\mathbf{x})}{t} =&\  \lim_{t\rightarrow 0} \frac{\frac{g(\mathbf{x}+t\mathbf{v})}{\|f(\mathbf{x}+t\mathbf{v})\|}-\frac{g(\mathbf{x})}{\|f(\mathbf{x})\|}}{t} = \lim_{t\rightarrow 0} \frac{g(\mathbf{x}+t\mathbf{v})\|f(\mathbf{x})\|-g(\mathbf{x})\|f(\mathbf{x}+t\mathbf{v})\|}{t\|f(\mathbf{x})\|\|f(\mathbf{x}+t\mathbf{v})\|}\\
=&\ \lim_{t\rightarrow 0} \frac{\left(g(\mathbf{x})+tdg(\mathbf{x})\cdot\mathbf{v}+\mathcal{O}(t^2)\right)\|f(\mathbf{x})\|-g(\mathbf{x})\left(\|f(\mathbf{x})\|+t\frac{f(\mathbf{x}) df(\mathbf{x})\cdot \mathbf{v}}{\|f(\mathbf{x})\|}+\mathcal{O}(t^2)\right)}{t\|f(\mathbf{x})\|\|f(\mathbf{x}+t\mathbf{v})\|}\\
=&\ \frac{\|f(\mathbf{x})\| dg(\mathbf{x}) - \frac{g(\mathbf{x})f(\mathbf{x})}{\|f(\mathbf{x})\|}df(\mathbf{x})}{\|f(\mathbf{x})\|^2}\cdot \mathbf{v} =: dh(\mathbf{x})\cdot \mathbf{v}.
\end{align}
Note: I have only used chain rule and one variable calculus. 
