How do you find the derivative of $y=arcsin(5x+5)$ ?

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Answer 1 · 2016-03-26T06:42:59

$(dy)/(dx)=5/sqrt(1-(5x+5)^2)$

If $u=arcsinv$ then $v=sinu$ , hence $(dv)/(du)=cosu$ and

$(du)/(dv)=1/cosu=1/sqrt(1-sin^2u)=1/sqrt(1-v^2)$

Hence, we can now find the the derivative of $y=arcsin(5x+5)$ using the concept function of a function and this is given by

$(dy)/(dx)=1/sqrt(1-(5x+5)^2)xxd/(dx)(5x+5)$ or

$(dy)/(dx)=5/sqrt(1-(5x+5)^2)$

How do you find the derivative of y=arcsin(5x+5)y=arcsin(5x+5)?