Derivative proof

**camjenson** · December 9th 2012, 11:12 PM

Show y=tan(arcsin(x)) then y'=((1-x^2))^-3/2

**Prove It** · December 9th 2012, 11:21 PM

Originally Posted by camjenson

Show y=tan(arcsin(x)) then y'=((1-x^2))^-3/2

$\displaystyle \begin{align*} y &= \tan{\left[ \arcsin{(x)} \right] } \\ &= \frac{\sin{\left[ \arcsin{(x)} \right]}}{\cos{\left[ \arcsin{(x)} \right]}} \\ &= \frac{x}{\sqrt{ 1 - \left\{ \sin{\left[ \arcsin{(x)} \right]} \right\}^2 }} \\ &= \frac{x}{\sqrt{1 - x^2}} \\ \\ \frac{dy}{dx} &= \frac{1 \, \sqrt{1 - x^2} - x \left( -\frac{x}{\sqrt{1 - x^2}} \right)}{1 - x^2} \\ &= \frac{\sqrt{1 - x^2} + \frac{x^2}{\sqrt{1 - x^2}}}{1 - x^2} \\ &= \frac{\frac{1 - x^2 + x^2}{\sqrt{1 - x^2}}}{1 - x^2} \\ &= \frac{1}{\left( 1 - x^2 \right) \sqrt{1 - x^2} } \\ &= \left( 1 - x^2 \right)^{-\frac{3}{2} } \end{align*}$

**MarkFL** · December 9th 2012, 11:25 PM

I would rewrite the function first before differentiating. Draw a right triangle with opposite side x and hypotenuse 1 to represent an angle whose sine is x. Now, use the Pythagorean theorem to find the adjacent side, then take the tangent of this angle. What do you get?

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