f(x)= 1/sq rt(x) + 2
find y '
is this answer -1 and if so if the sq root had x^2 under it would the answer be -2
$\displaystyle f(x) = \frac{1}{\sqrt x + 2}$
$\displaystyle f'(x) = \frac{(\sqrt x + 2)(0) - (1)(\frac{1}{2\sqrt x})}{(\sqrt x + 2)^2}$
$\displaystyle f'(x) = \frac{\frac{-1}{2\sqrt x}}{(\sqrt x + 2)^2}$
$\displaystyle f'(x) = -\frac{1}{2(\sqrt x)(\sqrt x + 2)^2}$
The derivative of $\displaystyle \sqrt x + 2$ is given by the power rule:
$\displaystyle \frac{d}{dx} x^n = nx^{n-1}$.
What you have is $\displaystyle x^{1/2} + 2$.
Taking the derivatives:
$\displaystyle \frac{d}{dx} x^{1/2} = (1/2)x^{-1/2} = \frac{1}{2\sqrt x}$
$\displaystyle \frac{d}{dx} 2 = 0$
$\displaystyle \frac{d}{dx} (\sqrt x + 2) = \frac{1}{2\sqrt x}$