Question

werewetrfg

Answer 1

i know how to do it without the definition but not sure how to with it

Answer 2

@zepdrix @.Sam. @iambatman @Compassionate @johnweldon1993

Answer 3

Oh the original? Always fun to go back to basics lol

\[\large f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}\]

so we have our function

\[\large f(x) = x + \sqrt{x}\]
so that means
\[\large f(x + h) = x + h + \sqrt{x + h}\]

so we have

\[\large f'(x) = \lim_{h \rightarrow 0} \frac{x + h + \sqrt{x + h} - (x + \sqrt{x})}{h}\]
\[\large f'(x) = \lim_{h \rightarrow 0} \frac{\cancel{x} + h + \sqrt{x + h} \cancel{- x} - \sqrt{x}}{h}\]
\[\large f'(x) = \lim_{h \rightarrow 0}\frac{h + \sqrt{x + h} -\sqrt{x}}{h}\]
Lets break all this up

\[\large f'(x) = \lim_{h \rightarrow 0}\frac{h}{h} + \frac{\sqrt{x + h}}{h} -\frac{\sqrt{x}}{h}\]
\[\large f'(x) = \lim_{h \rightarrow 0}1 + \frac{\sqrt{x + h} - \sqrt{x}}{h} \]

multiply by conjugate
\[\large f'(x) = \lim_{h \rightarrow 0}1 + \frac{\sqrt{x + h} - \sqrt{x}}{h} \times \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}} \]

\[\large f'(x) = \lim_{h \rightarrow 0}1 + \frac{\sqrt{x + h} - \sqrt{x}}{h} \times \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}} \]

after simplifying we have
\[\large f'(x) = \lim_{h \rightarrow 0}1 + \frac{\cancel{h} }{\cancel{h}\sqrt{x + h} + \cancel{h}\sqrt{x}} \]

\[\large f'(x) = \lim_{h \rightarrow 0}1 + \frac{1 }{\sqrt{x + h} + \sqrt{x}} \]

now it looks like we can let h = 0

\[\large f'(x) = \lim_{h \rightarrow 0}1 + \frac{1 }{\sqrt{x} + \sqrt{x}} \]

which is finally!!

\[\large f'(x) = 1 + \frac{1 }{2\sqrt{x}} \]

Answer 4

sorry it took a bit, lot to write out >.< lol

Answer 5

haha no problem. thank you so much