# Thread: Derivatives of inverse trigonometric functions

1. ## Derivatives of inverse trigonometric functions

Given that y=arcsinx, show that

(1-x^2)d^2y/dx^2 - x(dy/dx)=0

2. $\displaystyle \frac{dy}{dx}=\frac1{\sqrt{1-x^2}}$ whereat $\displaystyle \frac{d^2y}{dx^2}=\cdots$

Then just make the proper substitutions to prove it.

4. 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.

5. d2y/dx2 = (1-x^2)^-3/2...? I can't sub in dy/dx and d2y/dx2 and make it equal 0

6. $\displaystyle \frac{dy}{dx}=\frac{1}{\sqrt{1-x^{2}}}=\left( 1-x^{2} \right)^{-1/2}\implies \frac{d^{2}y}{dx^{2}}=-\frac{1}{2}\cdot (-2x)\left( 1-x^{2} \right)^{-3/2},$ whereat $\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{x}{\left( 1-x^{2} \right)^{3/2}}.$

Dinner is served.

7. OH man. I completely ignored the x in 2x. n00b error. Thank you for your help.

8. Hello, Erghhh!

It seems straight-forward enough.

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: . $\displaystyle (1-x^2)\;\;\cdot\;\;\frac{d^2y}{dx^2} \;\;-\;\;x\;\;\cdot\;\;\frac{dy}{dx}$

. . . $\displaystyle = \;(1-x^2)\cdot\frac{x}{(1-x^2)^{\frac{3}{2}}} - x\cdot\frac{1}{(1-x^2)^{\frac{1}{2}}}$

. . . $\displaystyle = \;\frac{x}{(1-x^2)^{\frac{1}{2}}} - \frac{x}{(1-x^2)^{\frac{1}{2}}}$

. . . $\displaystyle =\;0$

9. Thanks, i see what i'm doing now. i just made a stupid error.