Using the limit definition, how do you find the derivative of #f(x)=sqrt(2x-1)#?

1 Answer
Dec 17, 2015

#f'(x) = frac{ 1 }{ sqrt{2x-1} }#

Explanation:

#f'(x) = lim_{h->0} frac{f(x+h)-f(x)}{h}#

If the limit exists, this definition is called the First Principles of Derivatives .

#f'(x) = lim_{h->0} frac{f(x+h)-f(x)}{h}#

#= lim_{h->0} frac{sqrt{2(x+h)-1}-sqrt{2x-1}}{h}#

Multiply both the numerator and denominator by the conjugate surd, as #(sqrt{a}-sqrt{b})*(sqrt{a}+sqrt{b})=a-b#

#lim_{h->0} frac{sqrt{2x+2h-1}-sqrt{2x-1}}{h} = lim_{h->0} frac{ ( sqrt{2x+2h-1}-sqrt{2x-1} )( sqrt{2x+2h-1}+sqrt{2x-1} ) }{ h(sqrt{2x+2h-1}+sqrt{2x-1}) }#

#= lim_{h->0} frac{ (2x+2h-1)-(2x-1) }{ h(sqrt{2x+2h-1}+sqrt{2x-1}) }#

#= lim_{h->0} frac{ 2h }{ h(sqrt{2x+2h-1}+sqrt{2x-1}) }#

#= lim_{h->0} frac{ 2 }{ sqrt{2x+2h-1}+sqrt{2x-1} }#

#= frac{ 2 }{ sqrt{2x+2(0)-1}+sqrt{2x-1} }#

#= frac{ 2 }{ 2sqrt{2x-1} }#

#= frac{ 1 }{ sqrt{2x-1} }#