Question

differenciate using 1st principle square root of 2x f(x+h)-f(x)/h

Answer 1

i got 1/x^1/2 using power rule but not the same as when using 1st principle

Answer 2

is it that difficult to solve..

Answer 3

\[\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h \to 0}\frac{\sqrt{2(x+h)}-\sqrt{2x}}{h}\] is a start
then rationalize the numerator, cancel one factor of h and replace it by zero to take the limit

Answer 4

not working i got 1/squareroot of 2x..u gotta help pls

Answer 5

after rationalizing i got 2x-2h+2x/denominator which after taking limit becomes squareroot of 2x. and the numeratot 2x cancels -2x leaving

Answer 6

\[\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{2(x+h)}-\sqrt{2x}}{h}\]
\[=\frac{\sqrt{2x+2h}-\sqrt{2x}}{h}\]
\[=\frac{\sqrt{2x+2h}-\sqrt{2x}}{h}\times\frac{\sqrt{2x+2h}+\sqrt{2x}}{\sqrt{2x+2h}+\sqrt{2x}}\]
\[=\frac{2x+2h-2x}{h(\sqrt{2x+2h}+\sqrt{2x})}\]
\[=\frac{2h}{h(\sqrt{2x+2h}+\sqrt{2x})}\]
\[=\frac{2}{(\sqrt{2x+2h}+\sqrt{2x})}\]
now take the limit by replacing h by zero

Answer 7

thats 2/2x+squareroot of 2.. am i 2 solve futher

Answer 8

thats 1/xsquareroot of 2.

Answer 9

buh its not thesame as when using power rule which gives ue x^-1/2

Answer 10

thnks ill jus leave the answer like that thnks alot man..