Derivative of (x^2+x)^(1/2): Simple Solution Manual

[nameVoid]

D= (x^2+x)^(1/2)

I have a solutions manual here showing
D' = (1/2)(x^2+x)^(-1/2) (2x+1)(x')
but correct me if I am wrong..
D' = (1/2)(x^2+x)^(-1/2) (2x+1)(2x')

 

[rock.freak667]

well you didn't state what you were taking derivatives with respect to.

But if you have D(t)=f(x)

then D'(t)=f'(x)* (dx/dt) = f'(x)*x'

In your question; $f(x)=\sqrt{x^2+x}$

 

[nameVoid]

differentiating with respect to time t if you could please answer this simple question directly i would be glad especialy

 

[Mark44]

D= (x^2+x)^(1/2)

I have a solutions manual here showing
D' = (1/2)(x^2+x)^(-1/2) (2x+1)(x')
but correct me if I am wrong..
D' = (1/2)(x^2+x)^(-1/2) (2x+1)(2x')

The first one is correct; the second one is incorrect.