2.0 Some Basic Differentiation formulae

$$ y =f(x) = k$$ $$\frac{d}{dx}k = 0$$ Proof: $$f'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{k - k}}{h} = \mathop {\lim }\limits_{h \to 0} 0 = 0$$

Power rule, $$\frac{d}{dx}x^n = nx^{n-1}$$ Proof: For $n=1,$$$\begin{equation} \begin{aligned} y = x \\ \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {x + h} \right) - \left( x \right)}}{h} \\ \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \frac{h}{h} = \mathop {\lim }\limits_{h \to 0} 1 = 1 \\\end{aligned} \end{equation} $$ For $ n=2,$ $$\begin{equation} \begin{aligned} y = {x^2} \\ \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {x + h} \right)}^2} - \left( {{x^2}} \right)}}{h} \\\end{aligned} \end{equation} $$ Using Identity, $$\begin{equation} \begin{aligned} {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \\ \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{x^2} + {h^2} + 2xh} \right) - \left( {{x^2}} \right)}}{h} \\ \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{h^2} + 2xh} \right)}}{h} \\ \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {h + 2x} \right)}}{h} \\ \frac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \left( {h + 2x} \right) = 0 + 2x = 2x \\\end{aligned} \end{equation} $$ Thus, in general, $$\frac{d}{dx}x^n = nx^{n-1}$$