1. ## another derivative

How would you do
y = sin^-2 (x)/(1+x^2)
find yprime
I got an answer of (1+x^2)(-2sin^-1 (x) - 2xsin^-2 (x)/(1+x^2)^2

2. You say $\sin^{-2}x=\arcsin^2x$ isn't it?

3. Originally Posted by Krizalid
You say $\sin^{-2}x=\arcsin^2x$ isn't it?
Yeah i think those are equal.

4. Hello, davecs77!

Differentiate: . $y \:=\:\frac{(\sin x)^{-2}}{1+x^2}$

I would re-write it as: . $y \;=\;\frac{\csc^2\!x}{1+x^2}$

Then: . $y' \;=\;\frac{(1+x^2)\cdot2\csc x\cdot(-\csc x\cot x) - \csc^2\!x\cdot(2x)}{(1+x^2)^2}$

. . . . . $y' \;=\;\frac{-2(1+x^2)\csc^2\!x\cot x - 2x\csc^2\!x}{(1+x^2)^2}$

. . . . . $y' \;=\;\frac{-2\csc^2\!x\cdot\left[(1+x^2)\cot x - x\right]}{(1+x^2)^2}$