Question

derivative of
$\sqrt{x \:}$
by first principle​

Answer 1

Explanation + Answer:

$\dashrightarrow \: { \tt{f(x) = {x}^{ \frac{1}{2} } }} \\ \\ \dashrightarrow \: { \tt{f(x + h) = {(x + h)}^{ \frac{1}{2} } }}$

From first principles:

${ \rm{ \frac{ \delta y}{ \delta x} : { \rm{lim_{x \dashrightarrow 0} } \: \: \frac{f(x + h) - f(x)}{h} }}} \\ \\ { \rm{ \frac{ \delta y}{ \delta x} : { \rm{lim_{x \dashrightarrow 0} } \: \: { \rm{ \frac{(x + h) {}^{ \frac{1}{2} } - {x}^{ \frac{1}{2} } }{h} }}}}}$

• Rationalise the numerator:

${ \rm{ \frac{ \delta y}{ \delta x} : { \rm{lim_{x \dashrightarrow 0} } \: \: { \rm{ \frac{ { \{(x + h)}^{ \frac{1}{2} } - {x}^{ \frac{1}{2} } \} \{ {(x + h)}^{ \frac{1}{2} } + {x}^{ \frac{1}{2} } \}}{h \{ {(x + h)}^{ \frac{1}{2} } + {x}^{ \frac{1}{2} } \}} }}}}} \\ \\ { \rm{ \frac{ \delta y}{ \delta x} : { \rm{lim_{x \dashrightarrow 0} } \: \: { \rm{ \frac{(x + h) - x}{h \{ {(x + h)}^{ \frac{1}{2} } + {x}^{ \frac{1}{2} } \} } }}}}} \\ \\ { \rm{ \frac{ \delta y}{ \delta x} : { \rm{lim_{x \dashrightarrow 0} } \: \: { \rm{ \frac{h}{h}( \frac{1}{ {(x + h)}^{ \frac{1}{2} } + {x}^{ \frac{1}{2} } } ) }}}}} \\ \\ { \rm{ \frac{ \delta y}{ \delta x} : { \rm{lim_{x \dashrightarrow 0} } \: \: { \rm{ \: \: { \rm{ \frac{1}{ {(x + h)}^{ \frac{1}{2} } + {x}^{ \frac{1}{2} } } }}}}}}}$

• When h tends to zero, [ h → 0 ]

${ \boxed { \tt{ \frac{ \delta y}{ \delta x} \approx \frac{dy}{dx} }}}$

• Therefore;

${ \rm{ \frac{dy}{dx} = \frac{1}{ {(x + 0)}^{ \frac{1}{2} } + {x}^{ \frac{1}{2} } } }} \\ \\ { \rm{ \frac{dy}{dx} = \frac{1}{ {x}^{ \frac{1}{2} } + {x}^{ \frac{1}{2} } } }} \\ \\ { \rm{ \frac{dy}{dx} = \frac{1}{2 {x}^{ \frac{1}{2} } } }} \\ \\ { \boxed{ \boxed{ \sf{ \: \frac{dy}{dx} = \frac{1}{2 \sqrt{x} } }}}}$