Question

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

Answer 1

`(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)`