Question

Differentiate the following by using "limit"

Answer 1

Answer:

$\rm \displaystyle \frac{d }{dx} \sqrt{x} = \frac{ 1 }{ 2\sqrt{x } }$

Step-by-step explanation:

we want to differentiate the following by using limit:

$\displaystyle \frac{d}{dx} \sqrt{x}$

derivative definition by limit given by

$\rm \displaystyle \frac{d}{dx} = \lim _{\Delta x \to 0} \left( \frac{f(x + \Delta x) - f(x)}{ \Delta x} \right)$

given that,

f(x)=√x

so,

f(x+∆x)=√(x+∆x)

thus substitute:

$\rm \displaystyle \frac{d }{dx} \sqrt{x} = \lim _{\Delta x \to 0} \left( \frac{ \sqrt{x + \Delta x}- \sqrt{x} }{ \Delta x} \right)$

multiply both the numerator and denominator by the conjugate of the numerator:

$\rm \displaystyle \frac{d }{dx} \sqrt{x} = \lim _{\Delta x \to 0} \left( \frac{ \sqrt{x + \Delta x}- \sqrt{x} }{ \Delta x} \times \frac{ \sqrt{x + \Delta x} + \sqrt{x} }{\sqrt{x + \Delta x} + \sqrt{x}} \right)$

simplify which yields:

$\rm \displaystyle \frac{d }{dx} \sqrt{x} = \lim _{\Delta x \to 0} \left( \frac{ (\sqrt{x + \Delta x}) ^{2} - x }{ \Delta x(\sqrt{x + \Delta x} + \sqrt{x})} \right)$

simplify square:

$\rm \displaystyle \frac{d }{dx} \sqrt{x} = \lim _{ \Delta x \to 0} \left( \frac{ x + \Delta x - x }{ \Delta x(\sqrt{x + \Delta x} + \sqrt{x})} \right)$

collect like terms:

$\rm \displaystyle \frac{d }{dx} \sqrt{x} = \lim _{\Delta x \to 0} \left( \frac{ \Delta x }{ \Delta x(\sqrt{x + \Delta x} + \sqrt{x})} \right)$

reduce fraction:

$\rm \displaystyle \frac{d }{dx} \sqrt{x} = \lim _{\Delta x \to 0} \left( \frac{ 1 }{ (\sqrt{x + \Delta x} + \sqrt{x})} \right)$

get rid of ∆x as we are approaching its to 0

$\rm \displaystyle \frac{d }{dx} \sqrt{x} = \frac{ 1 }{ \sqrt{x } + \sqrt{x}}$

simplify addition:

$\rm \displaystyle \frac{d }{dx} \sqrt{x} = \frac{ 1 }{ 2\sqrt{x } }$

and we are done!