Question

find dy/dx if x^2 from the first principle​

Answer 1

First principle of derivative is nothing but,

$\displaystyle\mathsf{\dfrac {d}{dx}[f(x)]=\lim_{h\to 0}\dfrac {f(x+h)-f(x)}{h}}$

Here,

$\mathsf {f(x)=x^2}$

Then,

$\displaystyle\mathsf {\dfrac {d}{dx}[f(x)]=\lim_{h\to 0}\dfrac {f(x+h)-f(x)}{h}}\\\\\\\mathsf {\dfrac {d}{dx}(x^2)=\lim_{h\to 0}\dfrac {(x+h)^2-x^2}{h}}\\\\\\\mathsf {\dfrac {d}{dx}(x^2)=\lim_{h\to 0}\dfrac {x^2+2xh+h^2-x^2}{h}}\\\\\\\mathsf {\dfrac {d}{dx}(x^2)=\lim_{h\to 0}\dfrac {2xh+h^2}{h}}\\\\\\\mathsf {\dfrac {d}{dx}(x^2)=\lim_{h\to 0}\dfrac {h(2x+h)}{h}}\\\\\\\mathsf {\dfrac {d}{dx}(x^2)=\lim_{h\to 0}(2x+h)}\\\\\\\mathsf {\dfrac {d}{dx}(x^2)=2x+0}\\\\\\\mathsf {\dfrac {d}{dx}(x^2)=\underline {\underline {2x}}}$