Question

find the derivative of g(x)= arcsin 3x/x not really sure where to start i know that arcsin = U/Squ. rt. 1-u^2, but i dont understand how to use it correctly, please help

Answer 1

actually, \(\arcsin(u)=\dfrac{u'}{\sqrt{1-u^2}}\)

in your problem, \(u=3x\)

Answer 2

but fron start, you just use quotient rule: \(\dfrac{d}{dx}\dfrac{f(x)}{g(x)} = \dfrac{f'(x)~g(x)-f(x)~g'(x)}{[g(x)]^2}\)

you are given that \(\dfrac{f(x)}{g(x)} = \dfrac{\arcsin(3x)}{x}\) so \(f(x)=\arcsin(3x)\) and \(g(x)=x\)

Answer 3

Or you can just have \( \dfrac{\arcsin(3x)}{x}=x^{-1}\arcsin(3x)\) then use product rule from here

Answer 4

yeah i thought it could be used with quotient rule, but then i saw the other way