derivative?

Question

What is the second derivative of y=x*sqrt(16-x^2)?
i found the first derivative to be (-2x^2+16)/(sqrt(16-x^2))
is this correct?
and then i cnt do the frickin second derivative right!
how do you figure out the zeros of the second derivative (what are they when you figure out the second derivative?)
im so confused!! any help greatly appreciated. thanks!

Answer 1

Rewriting:
y = x(16-x^2)^1/2
dy/dx = [(16-x^2)^1/2]d/dx(x) + xd/dx[(16-x^2)^1/2] chain rule
y' = [(16-x^2)^1/2](1) + x[(1/2)(16-x^2)^(-1/2)](-2x)
= sqr(16-x^2) - 2(x^2)/sqr(16-x^2)
Try the second derivative using the same technique for the argument of the sqr. You can do it, just take as many steps as you need to keep the numbers straight.

Answer 2

oh god I hate derivatives, those are tripping me up so badly in IB calculus...but I'd recomment the quotient rule here (quotient rule works like this: if h(x)=f(x)/g(x), then hprime(x)=fprime(x)g(x)-gprime(x)f(x)/g(x)^2)

now, since the sqrt is essentially something to the 1/2 power, you'll have to add onto the numerator after the gprime(x)f(x), multiplying into that still, the derivative of the inner function, which is 16-x^2, and the derivative of the outer, which is (16-x^2)^1/2

oh, and where I say prime, that's a derivative. That should be how to get the second derivative....