Question

Solve

f(x)=1/(2x^(2/3))

using first principle of derivatives

Answer 1

\[f(x) = \frac{1}{2\sqrt[3]{x^2}}\]

Answer 2

\[\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h}\]\[\lim_{h \rightarrow 0}\frac{\frac{1}{2\sqrt[3]{(x+h)^2}}-\frac{1}{2\sqrt[3]{x^2}}}{h}\]

Answer 3

1/2x^-3/2=dy/dx=-3/4x^-5/2

Answer 4

\[\frac{\frac{2\sqrt[3]{x^2}-2\sqrt[3]{(x+h)^2}}{(2\sqrt[3]{(x+h)^2})(2\sqrt[3]{x^2})}}{h}\]

Answer 5

using first principle bnut

Answer 6

PLEASE SHOW ALL WORKING...

Answer 7

wait are you asking or telling us how to do it?

Answer 8

lol .... it doesnt matter that you own a john deere; go cut the yard with a pairs of rusty scissors :)

Answer 9

asking please :(...lol i want to know where i went wrong and what amistre :S lol

Answer 10

usually with someting like this; youwanna multiply top and bottom the conjugate of that top part

Answer 11

i did

Answer 12

i got -1/4x^(5/3) and i know its not right...

Answer 13

your common denom need not introduce in an extra "2"

Answer 14

\[\cfrac{\sqrt[3]{x^2}-\sqrt[3]{(x+h)^2}}{2h\sqrt[3]{x^2}\sqrt[3]{(x+h)^2}}\]

Answer 15

ok when i multiply byt the conjugate i get \[\frac{2x^2-2(x+h)^2}{(2\sqrt[3]{({x+h)^2)(2\sqrt[3]{x^2})(2\sqrt[3]{x^2}+2\sqrt[3]{(x+h)^2})}}}*\frac{1}{h}\]

Answer 16

how'd you get rid of the 2?

Answer 17

for simpicities sake:

\[\frac{1}{2(x+h)}*\frac{(x)}{(x)}-\frac{1}{2(x)}*\frac{(x+h)}{(x+h)}=\frac{x-(x+h)}{2x(x+h)}\]
your way is fine; it just puts in an extra "2" thats all

Answer 18

they have a 2 in common denoms already; so no need to include it again when combining them

Answer 19

oh okay lol...thats the really simple way but thanks i get it

Answer 20

:)

Answer 21

amistre i still dont know how to get the final answer though