Quotient rule with a square-root function

Let $f(x) = \sqrt{x}+1$ and $h(x) = \sqrt{x}-1$, so $g(x) = f(x)/h(x)$. Both $f'(x) = \frac{1}{2\sqrt{x}}$ and $h'(x) = \frac{1}{2\sqrt{x}}$ since the derivative of a constant is zero. By the quotient rule, $g'(x) = \frac{f'(x)h(x) - f(x)h'(x)}{h(x)^2} = \frac{\frac{1}{2\sqrt{x}}(\sqrt{x}-1) - (\sqrt{x}+1)\frac{1}{2\sqrt{x}}}{(\sqrt{x}-1)^2}.$

Factor $\frac{1}{2\sqrt{x}}$ from the numerator: $g'(x) = \frac{\frac{1}{2\sqrt{x}}\left[(\sqrt{x}-1)-(\sqrt{x}+1)\right]}{(\sqrt{x}-1)^2} = \frac{\frac{1}{2\sqrt{x}}(-2)}{(\sqrt{x}-1)^2} = \frac{-1}{\sqrt{x}(\sqrt{x}-1)^2}.$ Now substitute $x=4$, so $\sqrt{4}=2$: $g'(4) = \frac{-1}{2(2-1)^2} = -\frac{1}{2}.$

Negative. We found that g prime of four equals negative one half, which is a negative number. This follows directly from the simplified expression for g prime of x evaluated at x equals four.