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

Question

geerky42 · Answer

actually, $\arcsin(u)=\dfrac{u'}{\sqrt{1-u^2}}$

in your problem, $u=3x$

geerky42 · Answer

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 $g(x)=x$

geerky42 · Answer

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

anonymous · Answer

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