I was trying to derive the following result:
$\frac{df}{ds} = \frac{\partial f}{\partial x}\frac{dx}{ds}+\frac{\partial f}{\partial y}\frac{dy}{ds}$
For function $f(x(s),y(s))$. My derivation went like this:
$\frac{df}{ds}\\ =\lim_{h\rightarrow0} \frac{f(x(s+h),y(s+h))-f(x(s),y(s))}{h}\\ =\lim_{h\rightarrow0} \frac{f(x(s+h),y(s+h))-f(x(s),y(s+h))+f(x(s),y(s+h))-f(x(s),y(s))}{h}\\ =\lim_{h\rightarrow0}\frac{f(x(s+h),y(s+h))-f(x(s),y(s+h))}{h}+\lim_{h\rightarrow0}\frac{f(x(s),y(s+h))-f(x(s),y(s))}{h}\\ \approx\lim_{h\rightarrow0}\frac{f(x(s)+\delta x,y(s+h))-f(x(s),y(s+h))}{h}+\lim_{h\rightarrow0}\frac{f(x(s),y(s)+\delta y)-f(x(s),y(s))}{h}\\ =\lim_{h\rightarrow0}\frac{f(x(s)+\delta x,y(s+h))-f(x(s),y(s+h))}{\delta x}\lim_{h\rightarrow0}\frac{\delta x}{h}+\lim_{h\rightarrow0}\frac{f(x(s),y(s)+\delta y)-f(x(s),y(s))}{\delta y}\lim_{h\rightarrow0}\frac{\delta y}{h}$
So I'm not sure exactly how to proceed here. I understand that $\lim_{h\rightarrow0}\frac{f(x(s)+\delta x,y(s+h))-f(x(s),y(s+h))}{\delta x}$ will become $\frac{\partial f}{\partial x}$, but how exactly does that argument go? Something like:
$\lim_{h\rightarrow0}\frac{f(x(s)+\delta x,y(s+h))-f(x(s),y(s+h))}{\delta x}\\ \approx \lim_{\delta x\rightarrow0} \frac{f(x(s)+\delta x,y(s+h))-f(x(s),y(s+h))}{\delta x}\\ \approx \lim_{\delta x\rightarrow0} \frac{f(x(s)+\delta x,y(s))-f(x(s),y(s))}{\delta x}\\ =\frac{\partial f}{\partial x}$
Is the third line immediately above even correct? I know that the definition of $\frac{\partial f}{\partial x}$ is $\lim_{\delta x\rightarrow0} \frac{f(x(s)+\delta x,y(s))-f(x(s),y(s))}{\delta x}$ so I wouldn't think I can go straight from the second line to the fourth line, but I'm not sure and would appreciate some advice on how to correctly complete this derivation.
We will prove the following theorem, which will give you the desired result as a special case.
Theorem: Let $f$ be a scalar field defined on an open set $S$ in $\Bbb{R}^n$, and let $\mathbf{r}$ be a vector-valued function which maps an interval $J$ from $\Bbb{R}^1$ into $S$. Define the composite function $g=f\circ\mathbf{r}$ on $J$ by the equation $$g(s)=f[\mathbf{r}(s)], \quad s\in J.$$ Let $s$ be a point in $J$ at which $\mathbf{r}'(s)$ exists and assume that $f$ is differentiable at $\mathbf{r}(s)$. Then $g'(s)$ exists and is equal to the dot product $$ g'(s) = \nabla f(\mathbf{a})\cdot \mathbf{r}'(s), \quad \text{where}\quad \mathbf{a} = \mathbf{r}(s). $$ Proof: Let $\mathbf{a} = \mathbf{r}(s)$, where $s$ is a point in $J$ at which $\mathbf{r}'(s)$ exists. Since $S$ is open there is an $n$-ball $B(\mathbf{a})$ lying in $S$. We take $h \neq 0$ but small enough so that $\mathbf{r}(s+h)$ lies in $B(\mathbf{a})$, and we let $\mathbf{y} = \mathbf{r}(s+h)-\mathbf{r}(s)$. Note that $\mathbf{y}\to\mathbf{0}$ as $h \to 0$. Now we have $$g(s+h)-g(s)=f[\mathbf{r}(s+h)]-f[\mathbf{r}(s)]=f(\mathbf{a}+\mathbf{y})-f(\mathbf{a}).$$ Applying the first order Taylor formula for $f$ we have $$f(\mathbf{a}+\mathbf{y})-f(\mathbf{a})=\nabla f(\mathbf{a})\cdot\mathbf{y}+||\mathbf{y}||E(\mathbf{a},\mathbf{y}),$$ where $E(\mathbf{a},\mathbf{y})\to 0$ as $||\mathbf{y}||\to 0.$ Since $\mathbf{y}=\mathbf{r}(s+h)-\mathbf{r}(s)$ this gives us $$\frac{g(s+h)-g(s)}{h}=\nabla f(\mathbf{a})\cdot\frac{\mathbf{r}(s+h)-\mathbf{r}(s)}{h}+\frac{||\mathbf{r}(s+h)-\mathbf{r}(s)||}{h}E(\mathbf{a},\mathbf{y}).$$ Letting $h \to 0$ we obtain the required result.$\square$
Now coming to your special case we are working in $\Bbb{R}^2$, so we let $\mathbf{r}(s) = x(s)\mathbf{e_1}+y(s)\mathbf{e_2}$ where $\{\mathbf{e_1},\mathbf{e_2}\}$ are the standard basis vectors in $\Bbb{R}^2$. Applying the result of the theorem and writing in component form will give you the desired result.