What is the derivative of #(arcsin(3x))/x#?

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Answer 1 · 2015-04-23T22:33:57

You can solve this problem by using the quotient rule .

#y=(arcsin(3x))/x=u/v#

#u=arcsin(3x)#

#sinu=3x#

#cosu*(du)/(dx)=3#

#(du)/(dx)=3/cosu=3/sqrt(1-9x^2)#

#v=x#, #(dv)/(dx)=1#

#(dy)/(dx)=(x*3/sqrt(1-9x^2)-arcsin(3x)*1)/x^2#

#(dy)/(dx)=1/(x^2)*{(3x)/sqrt(1-9x^2)-arcsin(3x)}#