y= sqrt2x + 2sqrt2x
I need to find dy/dx,
When i differentiate each part I get the same result which I know is not correct,
y = 2x^-1/2 + 2(x^-1/2)
dy/dx = 1/2sqrtx + 1/2sqrtx
any help would be good
well using the power rule:
$\displaystyle f'(x)=nx^{n-1}$
We see in your problem:
$\displaystyle y= \sqrt{2x} + 2\sqrt{2x}$
1.
$\displaystyle y= (2x)^{1/2}+2(2x)^{1/2}$
2.
$\displaystyle
y= (x)^{-1/2}+2(x)^{-1/2}$
3.
$\displaystyle y= x^{-1/2}+2x^{-1/2}$
[quote=Zanderist;475964]well using the power rule:
$\displaystyle f'(x)=nx^{n-1}$
We see in your problem:
$\displaystyle y= \sqrt{2x} + 2\sqrt{2x}$
$\displaystyle y= (2x)^{1/2}+2(2x)^{1/2}$
$\displaystyle
y'= (x)^{-1/2}+2(x)^{-1/2}$ $\displaystyle \longleftarrow$ should be $\displaystyle y'= (2x)^{-1/2}+2(2x)^{-1/2}$
[\quote]
. . .
You had your help.
But it seems that if we comment on your sloppy use of language you take your balls back and run off. Now who is the loser, you want help, we give it (or rather DeDust, Halls, Zanderist and Drexel28 give it), but you won't play any more because I wish to correct a ridiculous use of language that I see all the time.
Ohhh.. well, my loss I suppose. Goodbye
CB