derrivatives of inverse trig functions

**johntuan** · October 14th 2008, 09:44 PM

G(x)= [sqrt(1-x^2)]arccosx

need help please.

**lllll** · October 14th 2008, 10:38 PM

$\frac{d}{dx} \ \sqrt{1-x^2}\arccos{x}$

$= \frac{1}{2}(1-x^2)^{-\frac{1}{2}} \times 2x \arccos{x} + \frac{-1}{\sqrt{1-x^2}} \times \sqrt{1-x^2}$

$= \frac{x}{\sqrt{1-x^2}} \arccos{x}-1$

**mr fantastic** · October 15th 2008, 12:43 AM

Originally Posted by lllll

$\frac{d}{dx} \ \sqrt{1-x^2}\arccos{x}$

$= \frac{1}{2}(1-x^2)^{-\frac{1}{2}} \times ({\color{red}-}2x) \arccos{x} + \frac{-1}{\sqrt{1-x^2}} \times \sqrt{1-x^2}$

$= \frac{{\color{red}-}x}{\sqrt{1-x^2}} \arccos{x}-1$

Small corrections in red.

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