Anti-Dif inverse circ functions

[TonyHem]

Yeah, just started this today so I suck.
How do I do:

"A curve has a gradient given by and its graph passes through the origin. Find the equation of the curve"

What I thought was x=0 since it passes through the origin, so dy/dx = 3. but it's wrong so yeah :{

[pHysiX]

They are asking for the original function, not the tangent to the curve or the gradient, which was what u did 
So do:

if dy/dx = that thing then
y=antidiff of 2 + antidiff of (1/SQRT(1-x^2)) +C

*I'm assuming that u can antidiff the first part. The second part: we know that the derivative of Sin^-1(x) is exactly "that", which implies the antidiff of "that" must be Sin^-1(x) =]

Since it passes the origin:

2x + Sin^-1(x) + C = y

y(0) ==> 0=C


Therefore:

y= 2x + Sin^-1(x)

[TonyHem]

zzzz -_-"
I get it now, thanks.