Question

Question #b6b90

Answer 1

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))`