Differentiate y = axⁿ using first principle method to differentiate.

The derivative $\frac{dy}{dx}$ is defined as:

$$\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

The binomial expansion of $(x+h)^n$ is $x^n + nx^{n-1}h + \frac{n(n-1)}{2!}x^{n-2}h^2 + \dots + h^n$.

$$\frac{dy}{dx} = \lim_{h \to 0} \frac{a \left( x^n + nx^{n-1}h + \frac{n(n-1)}{2!}x^{n-2}h^2 + \dots + h^n \right) - ax^n}{h}$$ $$\frac{dy}{dx} = \lim_{h \to 0} \frac{ax^n + anx^{n-1}h + a\frac{n(n-1)}{2!}x^{n-2}h^2 + \dots + ah^n - ax^n}{h}$$

The $ax^n$ terms cancel out:

$$\frac{dy}{dx} = \lim_{h \to 0} \frac{anx^{n-1}h + a\frac{n(n-1)}{2!}x^{n-2}h^2 + \dots + ah^n}{h}$$

Divide each term in the numerator by $h$:

$$\frac{dy}{dx} = \lim_{h \to 0} \left( anx^{n-1} + a\frac{n(n-1)}{2!}x^{n-2}h + \dots + ah^{n-1} \right)$$

As $h \to 0$, all terms containing $h$ will become zero, except for the first term $anx^{n-1}$.

$$\frac{dy}{dx} = anx^{n-1}$$

The derivative of $y = ax^n$ using the first principle method is: