This is what I tried:
$$ {1 \over {1 + \sqrt x }} = {1 \over {\sqrt {1 + x} }} = {1 \over {{{(1 + x)}^{{1 \over 2}}}}} = {(1 + x)^{ – {1 \over 2}}} $$
$${{dy} \over {dx}} = – {1 \over 2}{(1 + x)^{ – {3 \over 2}}}(1) = {{ – 1} \over {2{{(1 + x)}^{{3 \over 2}}}}}$$
This is the wrong answer, I'm certain I've made a mistake in interpreting or applying the chain rule.
If anyone could point out where I've went wrong in my approach that would be great, thank you.
Best Answer
Your problem was not in the application of the chain rule; the problem is in the following:
That is, $$\frac{1}{1 + \sqrt x} \neq \frac{1}{(1+x)^{1\over 2}}$$
Since you used an non-equivalent function, although your application of the chain-rule on the incorrect function was correct, it was not the derivative of the given function.
If we let $\color{green}{\bf \;u = (1 + \sqrt x)},\;$ then $\;\color{blue}{\bf\dfrac{du}{dx} = \dfrac{1}{2\sqrt x}}$.
Then we have the function $\quad y = \dfrac 1u = u^{-1}.\;$ $\color{red}{\bf \dfrac{dy}{du}} = -u^{-2} = \color{red}{\bf \dfrac{-1}{ u^2}}.\;$ The derivative of the function is $$\frac{dy}{dx} = \color{red}{\bf \frac{dy}{du}} \cdot \color{blue}{\bf \dfrac{du}{dx}} = \color{red}{\bf \frac{-1}{u^{2}}}\cdot \color{blue}{\bf \frac{1}{2\sqrt x}} = \dfrac{-1}{\color{green}{\bf(1+\sqrt x)}^2}\cdot \dfrac{1}{2\sqrt x} = \dfrac{-1}{2\sqrt x(1+\sqrt x)^2}$$