B bobers New member Joined Nov 29, 2008 Messages 10 Dec 3, 2008 #1 Derivative of inverse functions help? How do you show that that the function f(x)=arcsin((x-2)/2)-2arcsin((x^1/2)/2... is constant for 0?x?4 i took the derivative and put 0 and 4 and got 0 But numbers between the two like 2 doesn't work
Derivative of inverse functions help? How do you show that that the function f(x)=arcsin((x-2)/2)-2arcsin((x^1/2)/2... is constant for 0?x?4 i took the derivative and put 0 and 4 and got 0 But numbers between the two like 2 doesn't work
skeeter Elite Member Joined Dec 15, 2005 Messages 3,204 Dec 3, 2008 #2 \(\displaystyle f(x) = \arcsin(u) - 2\arcsin(v)\) \(\displaystyle u = \frac{x-2}{2}\) \(\displaystyle u^2 = \frac{(x-2)^2}{4}\) \(\displaystyle u' = \frac{1}{2}\) \(\displaystyle v = \frac{\sqrt{x}}{2}\) \(\displaystyle v^2 = \frac{x}{4}\) \(\displaystyle v' = \frac{1}{4\sqrt{x}}\) \(\displaystyle f'(x) = \frac{u'}{\sqrt{1 - u^2}} - \frac{2v'}{\sqrt{1 - v^2}}\) sub in the values and do the grunt work algebra ... you should get \(\displaystyle f'(x) = 0\)
\(\displaystyle f(x) = \arcsin(u) - 2\arcsin(v)\) \(\displaystyle u = \frac{x-2}{2}\) \(\displaystyle u^2 = \frac{(x-2)^2}{4}\) \(\displaystyle u' = \frac{1}{2}\) \(\displaystyle v = \frac{\sqrt{x}}{2}\) \(\displaystyle v^2 = \frac{x}{4}\) \(\displaystyle v' = \frac{1}{4\sqrt{x}}\) \(\displaystyle f'(x) = \frac{u'}{\sqrt{1 - u^2}} - \frac{2v'}{\sqrt{1 - v^2}}\) sub in the values and do the grunt work algebra ... you should get \(\displaystyle f'(x) = 0\)