Question #b6b90

1 Answer
Nov 2, 2017

y'=arcsenx+x/(√(1-x^2))+1/(√x(2x+2))

Explanation:

Use product rule
y=xarcsenx
y'=x'(arcsenx)+x(arcsenx)'

Derivative of x=1
Derivative of arcsenx= 1/(√(1-x^2))
Then
y'=arcsenx+x/(√(1-x^2))

Now for arctg√x

derivative of arctg(u) where u is a function

Is (d/dx(u))/(1+u^2)
Then for √x

Derivative of √x is 1/(2√x)
Then
Derivative of arctg√x= (1/(2√x))/(1+x)

-> 1/(√x(2x+2))