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

1 Answer
Shwetank Mauria
Mar 26, 2016

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

Explanation:

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)#

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