Derivative Help

[davidiswhat]

dy/dx = (1-y^2)^1/2
find d^2y/dx^2
A: -y
so to get it ...
1/2(1-y^2)^-1/2(-2y) //chain rule ... dont get why a dy/dx should be on top
-2y/2dy/dx
So to get the answer im suppose to cancel the denominator by dy/dx and i wish to know how/why we got that. i kinda get that we're solving y'(y) instead of y'(x) but im sill confuse. Also i think we used to do something like dx/dx when we did y'(x) so if it was a x instead ... y''(x)=1/2(1-x^2)^-1/2(-2x) dx/dx why is this here?
basically
-y(dy/dx) <- ?
/dy/dx

 

[HallsofIvy]

You have "chain rule" in your answer- do you know what that is? The chain rule says that if we have f(y), a function of variable y, and y itself is a function of x, then \(\displaystyle \frac{df}{dx}= \frac{df}{dy}\frac{dy}{dx}\).

Here, \(\displaystyle f(y)= (1- y^2)^{1/2}\) so \(\displaystyle \frac{df}{dy}= (1/2)(1- y^2)^{-1/2}(-2y)= y(1- y^2)^{-1/2}\) just as you say. And you used the chain rule to get that 2y- that's probably what you meant. But now, to get the derivative with respect to x, you need to multiply by \(\displaystyle \frac{dy}{dx}\): \(\displaystyle \frac{d^2y}{dx^2}= \frac{d\left)\frac{dy}{dx}\right)}{dy}\frac{dy}{dx}= \left(y(1- y^2)^{-1/2}\right)(1- y^2)^{-1/2}\).

 

[davidiswhat]

Here, \(\displaystyle f(y)= (1- y^2)^{1/2}\) so \(\displaystyle \frac{df}{dy}= (1/2)(1- y^2)^{-1/2}(-2y)= y(1- y^2)^{-1/2}\) just as you say.
[/tex]
But now, to get the derivative with respect to x, you need to multiply by \(\displaystyle \frac{dy}{dx}\): \(\displaystyle \frac{d^2y}{dx^2}= \frac{d\left)\frac{dy}{dx}\right)}{dy}\frac{dy}{dx}= \left(y(1- y^2)^{-1/2}\right)(1- y^2)^{-1/2}\).[/QUOTE]

-y(1- y^2)^{-1/2} right ...
i think i get it d(dy/dx)/dy = df/dy and then you multiple that with dy/dx but
like what about when dy/dx is not given like
f(y(x)) = 3y(x)
f=3y
df/dx=(df/dy)(dy/dx)
so f ' = 3 (dy/dx)?
df/dy = 3 and dy/dx cant be solved?