What is the derivative of #arcsin[x^(1/2)]#?

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Answer 1 · 2015-06-06T04:04:13

To find the derivative we will need to use the Chain Rule

#dy/dx=dy/(du)*(du)/(dx)#

We want to find

#d/(dx)(arcsin(x^(1/2)))#

Following the chain rule we let #u=x^(1/2)#

Deriving u we get

#(du)/(dx)=1/2*x^(-1/2)=1/(2sqrt(x))#

Now we substitute u in place of x in the original equation and derive to find #dy/(du)#

#y=arcsin(u)#

#(dy)/(du)=1/(sqrt(1-u^2)#

Now we substitute these derived values into the chain rule to
find #dy/(dx)#

#dy/dx=dy/(du)*(du)/(dx)#

#dy/dx=1/(sqrt(1-u^2))*1/(2sqrt(x))#

Substitute x back into the equation to get the derivative in terms of x only and simplify

#u=x^(1/2)#

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

#dy/(dx)=1/(sqrt(1-x))*1/(2sqrt(x))#

#dy/(dx)=1/(2sqrt(x)*sqrt(1-x))#

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