Question

find the derivative of the following function from first principle f(x)= (x+1)/(x+2)
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Answer 1

EXPLANATION.

Derivative of the function from first principal.

→ f(x) = ( x + 1 ) / ( x + 2 )

$\sf : \implies \: let \: f(x) = \dfrac{(x + 1)}{(x + 2)} \\ \\ \sf : \implies \: by \: using \: first \: principal \\ \\ \sf : \implies \: \: f'(x) = \lim_{h \to \: 0} \: \frac{f(x + h) - f(x)6}{h}$

$\sf : \implies \: \lim_{h \: \to \: 0} \: = \dfrac{ \dfrac{x + h + 1}{x + h + 2} \: \: - \: \: \dfrac{x + 1}{x + 2} }{h} \\ \\ \sf : \implies \: \lim_{h \: \to \: 0} \: \frac{1}{h} [ \frac{(x + h + 1)(x + 2) - (x + 1)(x + h + 2)}{(x + h + 2)(x + 2)} ] \\ \\ \sf : \implies \: \lim_{h \: \to \: 0} \: = \frac{1}{h} [ \frac{( {x}^{2} + 2x + hx + 2h + x + 2) - ( {x}^{2} + hx + 2x + x + h + 2 }{(x + h + 2)(x + 2)} ]$

$\sf : \implies \: \lim_{h \: \to \: 0} \: = \dfrac{1}{h} [ \dfrac{ {x}^{2} + 3x + hx + 2h + 2 - {x}^{2} - hx - 3x - h - 2 }{(x + h + 2)(x + 2)} ] \\ \\ \sf : \implies \: \lim_{h \: \to \: 0} \: = \frac{1}{h} [ \frac{2h - h}{(x + h + 2)(x + 2)} ] \\ \\ \sf : \implies \: \lim_{h \: \to \: 0} \: = \frac{1}{ \cancel{h}} [ \frac{ \cancel{h}}{(x + h + 2)(x + 2)} ] \\ \\ \sf : \implies \: \lim_{h \to \: 0} \: = \frac{1}{(x + h + 2)(x + 2)}$

$\sf : \implies \: put \: the \: value \: of \: h \: = 0 \\ \\ \sf : \implies \: \lim_{h \: \to \: 0} = \frac{1}{(x + 0 + 2)(x + 2)} \\ \\ \sf : \implies \: \frac{1}{(x + 2) {}^{2} } = answer$