Question

If f ( x ) = x² , Find f ' ( x ) using First Principal of Differentiation​

Answer 1

Solution :-

First principal of Differentiation states that for any function f(x), it's differentiation would be:

$\boxed{ \red{ \boldsymbol{f'(x) = \lim_{h \to 0} \: \frac{f(x + h) - f(x)}{h} }}}$

According to the question,

• f (x) = x²

→ So, f (x + h) = (x + h)²

Now,

$\implies \sf{f'(x) = \lim_{h \to 0} \: \dfrac{(x + h)^{2} - {x}^{2} }{h} }$

Split (x + h)² using identity (a + b)² = a² + b² + 2ab

$\implies \sf{f'(x) = \lim_{h \to 0} \: \dfrac{ {x}^{2} + {h}^{2} + 2xh - {x}^{2} }{h} }$

$\implies \sf{f'(x) = \lim_{h \to 0} \: \dfrac{ \cancel{{x}^{2}} + {h}^{2} + 2xh - \cancel{{x}^{2}} }{h} }$

$\implies \sf{f'(x) = \lim_{h \to 0} \: \: \dfrac{ {h}^{2} + 2xh }{h} }$

$\implies \sf{f'(x) = \lim_{h \to 0} \: \: \dfrac{ \not{h}(h + 2x)}{ \not{h}} }$

$\implies \sf{f'(x) = \lim_{h \to 0} \: \: (h + 2x)}$

Substituting the limits,

$\implies \sf{f'(x) = (0 + 2x)}$

$\therefore \: \boxed{ \pink{ \bf{f'(x) = 2x}}}$