Question

why is the derivative of y=sqrt(x), y'=1/2x^-1/2?

Answer 1

are you asking how to get this from scratch?

\[\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\]?

Answer 2

if you know the power rule it is a straightforward application

Answer 3

yea that is what im asking from scratch like why ise power the same still?

Answer 4

why is the power the same still?*

Answer 5

sqrt(x) = x^1/2

Answer 6

well power rule says
\[\frac{d}{dx}x^r=rx^{r-1}\]

Answer 7

1 * 1/2 = 1/2

Answer 8

and
\[\sqrt{x}=x^{\frac{1}{2}}\]

Answer 9

1x^1/2

Answer 10

so power rule give it to you right away

Answer 11

\[\color{red}{\text{right myininaya?}}\]

Answer 12

power rules of x^n = nx^(n-1)

Answer 13

yea i have to subtract by one..i did it on the calculator and got the answer

Answer 14

but you can always compute
\[\lim_{\rightarrow 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\] and get the same thing

Answer 15

yeah, but it simplifies to d/dx x^n = nx^(n-1)

Answer 16

btw this function is very common, so it is best just to remember it and not use the power rule each time. kind of like memorizing what 7 times 8 is

Answer 17

\[\lim_{h \rightarrow 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}=\lim_{h \rightarrow 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}*\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\]
\[=\lim_{h \rightarrow 0}\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x}}=\lim_{h \rightarrow 0}\frac{1}{\sqrt{x+h}+\sqrt{x}}=\frac{1}{\sqrt{x+0}+\sqrt{x}}\]
\[=\frac{1}{2\sqrt{x}}\]

Answer 18

yes, but proving it to yourself first makes you feel confident about it

Answer 19

what myininaya said. but also if you are taking calc, you should just remember that the derivative of the square root of x is one over 2 root x. it will help you for sure

Answer 20

oops i forgot to close one of my parenthesis :(

Answer 21

ohhh okay bt when i need to show my work the power rule will be in memory

Answer 22

also, remember x^(-1/2) = 1/(x^(1/2))