Given that y=arcsinx, show that
(1-x^2)d^2y/dx^2 - x(dy/dx)=0
You're expected to know that $\displaystyle \frac{d^2y}{dx^2}$ is the second derivative, whereas the first is $\displaystyle \frac1{\sqrt{1-x^2}}$ then by just knowing this you can compute $\displaystyle \frac{d^2y}{dx^2}.$ Once you get $\displaystyle \frac{dy}{dx}$ and $\displaystyle \frac{d^2y}{dx^2}$ put them together into the equation and the rest is algebra.
$\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1-x^{2}}}=\left( 1-x^{2} \right)^{-1/2}\implies \frac{d^{2}y}{dx^{2}}=-\frac{1}{2}\cdot (-2x)\left( 1-x^{2} \right)^{-3/2},$ whereat $\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{x}{\left( 1-x^{2} \right)^{3/2}}.$
Dinner is served.
Hello, Erghhh!
It seems straight-forward enough.
Exactly where is your difficulty?
Given that: .$\displaystyle y\:=\:\arcsin x$,
show that: .$\displaystyle (1-x^2)\,\frac{d^2y}{dx^2} - x\,\frac{dy}{dx} \:=\:0$
We have: .$\displaystyle \frac{dy}{dx} \:=\:\frac{1}{\sqrt{1-x^2}} \:=\:(1-x^2)^{-\frac{1}{2}} $
And: .$\displaystyle \frac{d^2y}{dx^2} \;=\;-\frac{1}{2}(1-x^2)^{-\frac{3}{2}} (-2x) \;=\;\frac{x}{(1-x^2)^{\frac{3}{2}}} $
Then: . $\displaystyle (1-x^2)\;\;\cdot\;\;\frac{d^2y}{dx^2} \;\;-\;\;x\;\;\cdot\;\;\frac{dy}{dx} $
. . . $\displaystyle = \;(1-x^2)\cdot\frac{x}{(1-x^2)^{\frac{3}{2}}} - x\cdot\frac{1}{(1-x^2)^{\frac{1}{2}}} $
. . . $\displaystyle = \;\frac{x}{(1-x^2)^{\frac{1}{2}}} - \frac{x}{(1-x^2)^{\frac{1}{2}}}$
. . . $\displaystyle =\;0 $