Answer:
[tex] \rm \displaystyle \frac{d }{dx} \sqrt{x} = \frac{ 1 }{ 2\sqrt{x } } [/tex]
Step-by-step explanation:
we want to differentiate the following by using limit:
[tex] \displaystyle \frac{d}{dx} \sqrt{x} [/tex]
derivative definition by limit given by
[tex] \rm \displaystyle \frac{d}{dx} = \lim _{\Delta x \to 0} \left( \frac{f(x + \Delta x) - f(x)}{ \Delta x} \right) [/tex]
given that,
f(x)=√x
so,
f(x+∆x)=√(x+∆x)
thus substitute:
[tex] \rm \displaystyle \frac{d }{dx} \sqrt{x} = \lim _{\Delta x \to 0} \left( \frac{ \sqrt{x + \Delta x}- \sqrt{x} }{ \Delta x} \right) [/tex]
multiply both the numerator and denominator by the conjugate of the numerator:
[tex] \rm \displaystyle \frac{d }{dx} \sqrt{x} = \lim _{\Delta x \to 0} \left( \frac{ \sqrt{x + \Delta x}- \sqrt{x} }{ \Delta x} \times \frac{ \sqrt{x + \Delta x} + \sqrt{x} }{\sqrt{x + \Delta x} + \sqrt{x}} \right) [/tex]
simplify which yields:
[tex] \rm \displaystyle \frac{d }{dx} \sqrt{x} = \lim _{\Delta x \to 0} \left( \frac{ (\sqrt{x + \Delta x}) ^{2} - x }{ \Delta x(\sqrt{x + \Delta x} + \sqrt{x})} \right) [/tex]
simplify square:
[tex] \rm \displaystyle \frac{d }{dx} \sqrt{x} = \lim _{ \Delta x \to 0} \left( \frac{ x + \Delta x - x }{ \Delta x(\sqrt{x + \Delta x} + \sqrt{x})} \right) [/tex]
collect like terms:
[tex] \rm \displaystyle \frac{d }{dx} \sqrt{x} = \lim _{\Delta x \to 0} \left( \frac{ \Delta x }{ \Delta x(\sqrt{x + \Delta x} + \sqrt{x})} \right) [/tex]
reduce fraction:
[tex] \rm \displaystyle \frac{d }{dx} \sqrt{x} = \lim _{\Delta x \to 0} \left( \frac{ 1 }{ (\sqrt{x + \Delta x} + \sqrt{x})} \right) [/tex]
get rid of ∆x as we are approaching its to 0
[tex] \rm \displaystyle \frac{d }{dx} \sqrt{x} = \frac{ 1 }{ \sqrt{x } + \sqrt{x}} [/tex]
simplify addition:
[tex] \rm \displaystyle \frac{d }{dx} \sqrt{x} = \frac{ 1 }{ 2\sqrt{x } } [/tex]
and we are done!
