derivative of inverse trig: show f(x)=arcsin((x-2)/2)-2arcsi

[bobers]

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]

$\displaystyle f(x) = \arcsin(u) - 2\arcsin(v)$

$\displaystyle u = \frac{x-2}{2}$

$\displaystyle u^2 = \frac{(x-2)^2}{4}$

$\displaystyle u' = \frac{1}{2}$

$\displaystyle v = \frac{\sqrt{x}}{2}$

$\displaystyle v^2 = \frac{x}{4}$

$\displaystyle v' = \frac{1}{4\sqrt{x}}$

$\displaystyle f'(x) = \frac{u'}{\sqrt{1 - u^2}} - \frac{2v'}{\sqrt{1 - v^2}}$

sub in the values and do the grunt work algebra ... you should get $\displaystyle f'(x) = 0$