Deriving the derivative of $(1+x^m)^n$ from first principles and the binomial expansion? From the chain rule:
$$\frac{d}{dx} (1+x^m)^n = nmx^{m-1}(1+x^m)^{n-1}$$
And I am trying to prove an this result using only the first two terms of the binomial expansion and first principles.

Here is my attempt:
$$(1+x^m)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$$
Keeping just the first two terms and using differentiation by first principles:
$$
\begin{align}
\frac{d}{dx} (1 + x^m)^n &= \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{(1 + (x + \Delta x)^m)^n - (1 + x^m)^n}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{(1 + x^m(1 + \frac{\Delta x}{x})^m)^n - (1 + nx^m)}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{(1 + x^m(1 + \frac{\Delta x}{x})^m)^n - (1 + nx^m)}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{(1 + x^m(1 + m \frac{\Delta x}{x}))^n - (1 + nx^m)}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{(1 + nx^m(1 + m \frac{\Delta x}{x})) - (1 + nx^m)}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{nx^m(m \frac{\Delta x}{x}))}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{nmx^{m-1}\Delta x}{\Delta x} \\
&= nmx^{m-1} \\
\end{align}
$$
 A: It should be
$$(1+x^m)^n = 1 + nx^{\color{red}{m}} + \frac{n(n-1)}{2!}x^{\color{red}{2m}} + \dots+nx^{m(n-1)}+x^{mn}$$
then if you are using only the first term, you need to show that the remaining terms are of order higher than $\Delta x$, and that the lower order terms cancel. So we have
$$\frac{(1+(x+\Delta x)^m)^n-(1+x^m)^n}{\Delta x}$$
$$=\frac{1+n(x+\Delta x)^m+\dots+n(x+\Delta x)^{m(n-1)}+(x+\Delta x)^{mn}-1-nx^{mn}-\cdots-x^{mn}}{\Delta x}$$
$$=\frac{1+nx^{m}(1+\frac{\Delta x}{x})^{m}+\cdots+x^{mn}(1+\frac{\Delta x}{x})^{mn}-1-nx^{mn}-\dots-x^{mn}}{\Delta x}\quad(1)$$
$$=\frac{\Delta x\left(mnx^{m-1}+mx^{m-1}n(n-1)x^{m}+\cdots+mnx^{m-1}x^{m(n-1)}\right)+O((\Delta x)^2)}{\Delta x}$$
$$\rightarrow mnx^{m-1}\left(1+(n-1)x^{m}+\frac{(n-1)(n-2)}{2!}x^{2m}+...+x^{m(n-1)}\right)=mnx^{m-1}(1+x^m)^{n-1}$$
as $\Delta x\rightarrow 0$,  where the the first term of each bracket in the numerator of $(1)$ cancels with the negative terms, after applying the binomial expansion again.
A: For $h\to 0$, $$\frac{(1+(x+h)^m)^n -(1+x^m)^n}{h} \\ = \frac{\left(1+x^m(1+\frac hx)^m \right)^n-(1+x^m)^n}{h} \\ \to \frac{\left(1+x^m(1+\frac{mh}{x}) \right)-(1+x^m)^n}{h} \\ = \frac{(1+x^m +mhx^{m-1})^n -(1+x^m)^n}{h} \\ = \frac 1h \sum_{r=1}^{\infty} \binom nr (mhx^{m-1})^r(1+x^m)^{n-r} \\ \to  nmx^{m-1} (1+x^m)^{n-1}$$
as required.
