Question

f(x) = x+x^(1/2)
find the derivative using the definition of derivative

Answer 1

whatcha got so far?

Answer 2

ummm i know that it's lim of h approaching zero of the equation f(x+h) - f(x) all over h

Answer 3

well, thats a start :)

Answer 4

but i cant get the h out of the denominator

Answer 5

the root is going to be a pain in the neck
have fun!

Answer 6

do you know that the derivative of a sum or difference, is the difference of the derivatives of the parts?

Answer 7

g(x) = x ; g(x+h) = x+h
\[\lim_{h->0}\frac{g(x+h)-g(x)}{h}\]
\[\lim_{h->0}\frac{x+h-x}{h}\]
\[\lim_{h->0}\frac{h}{h}\]
\[\lim_{h->0}\ 1=1\]

Answer 8

s(x) = sqrt(x) ; s(x+h) = sqrt(x+h)
\[\lim_{h->0}\frac{s(x+h)-s(x)}{h}\]
\[\lim_{h->0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\]
\[\lim_{h->0}\frac{\sqrt{x+h}-\sqrt{x}}{h}*\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\]
\[\lim_{h->0}\ \frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}\]
\[\lim_{h->0}\ \frac{h}{h(\sqrt{x+h}+\sqrt{x})}\]
\[\lim_{h->0}\ \frac{1}{\sqrt{x+h}+\sqrt{x}}\]
\[\lim_{h->0}\ \frac{1}{\sqrt{x+0}+\sqrt{x}}=\frac{1}{2\sqrt{x }}\]

Answer 9

sooo..... out them together now :)

Answer 10

or put them together .... you can out them if you want, but it just doesnt have the desired effect

Answer 11

very nice! btw @casey this derivative comes up so frequently you should just remember it. it will save you a ton of time later. like memorizing 8*7