2nd derivative of cos^-1(x)

[M TI]

Hello,

I was wondering if someone could help me. The derivative of inverse cosine(x) is (acos(x))′ = \(\displaystyle -\, \dfrac{1}{\sqrt{\strut 1\, -\, x^2\,}}\)

I need to find the 2nd derivative of this function but I'm struggling where to start, how to break it down, etc. It's been a while and I am quite rusty!! Please, could someone help me with this explaining in detail the steps to get to the 2nd derivative?

regards

Matt.

 

[ksdhart2]

I can think of two ways you might find the derivative. The first involves the chain rule and the second involves the Quotient Rule. For the first, we can manipulate the equation to make a bit more obvious what's going on:

\(\displaystyle \displaystyle -\frac{1}{\sqrt{1-x^2}}=\left(-\sqrt{1-x^2}\right)^{-1}=-\left(1-x^2\right)^{-\frac{1}{2}}\)

So to use the chain rule, let f(x) = x^-1/2 and g(x) = 1 - x². The quotient rule doesn't require manipulating the equation at all. There, f(x) = 1 and g(x) = sqrt(1 - x²)

EDIT: Yup, I lost a minus sign. Whoops. I guess I get Five for Fighting.

 

[Steven G]

I can think of two ways you might find the derivative. The first involves the chain rule and the second involves the Quotient Rule. For the first, we can manipulate the equation to make a bit more obvious what's going on:

\(\displaystyle \displaystyle -\frac{1}{\sqrt{1-x^2}}=\left(-\sqrt{1-x^2}\right)^{-1}=\left(1-x^2\right)^{-\frac{1}{2}}\)

So to use the chain rule, let f(x) = x^-1/2 and g(x) = 1 - x². The quotient rule doesn't require manipulating the equation at all. There, f(x) = 1 and g(x) = sqrt(1 - x²)

You lost a '-' sign. 2 minutes in the penalty box??