Find dy/dx given $y = {1 \over {1 + \sqrt x }}$, using the chain rule 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.
 A: $$1+\sqrt{x} \ne \sqrt{1+x}$$
Rather,
$$\frac{d}{dx} \frac{1}{1+\sqrt{x}} = -\frac{1}{(1+\sqrt{x})^2} \frac{d}{dx}(1+\sqrt{x})$$
A: Well, let's go step by step. We are given the function $f: \mathbb{R} \to \mathbb{R}$ given by:
$$f(x)=\frac{1}{1+\sqrt{x}}$$
The chain rule is used to differentiate compositions. So let's write this as a composition? Well, i'll define the function $g: \mathbb{R}^+ \to \mathbb{R}$ by:
$$g(x)=\frac{1}{1+x}$$
And define the function $h : \mathbb{R} \to \mathbb{R}$ by $h(x) = \sqrt{x}$. It turns out pretty trivially that we must have $f = g \circ h$ because:
$$f(x)=g(h(x))=\frac{1}{1+\sqrt{x}}$$
Now, it's just application of the chain rule, we get the following:
$$f'(x)=g'(h(x))h'(x)$$
Now it's just computation, you know that $g'(h(x))$ is just about differentiating $g$ and applying at $h(x)$, so that we get:
$$g'(h(x))=\frac{-1}{(1+\sqrt{x})^2}$$
$$h'(x)=\frac{1}{2\sqrt{x}}$$
So the derivative of $f$ at $x$ must be:
$$f'(x)=\frac{-1}{2\sqrt{x}(1+\sqrt{x})^2}$$
Now, this kind of thing, making everything explicit, defining each function, writing down the derivatives, and writing down the composition, is something that we only do in a few examples to understand what's the idea behind. Once you understood, do it on your mind. I hope this detailed example helps you out.
A: Your problem was not in the application of the chain rule; the problem is in the following:

$\eqalign{
 {1 \over {1 + \sqrt x }} \quad \overset{\color{red}{\bf \Large \checkmark }}= \quad {1 \over {\sqrt {1 + x} }} = {1 \over {{{(1 + x)}^{{1 \over 2}}}}}  \cr }$

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 $$\frac{dy}{dx} = \color{red}{\bf \frac{dy}{du}} \cdot \color{blue}{\bf \dfrac{du}{dx}} = \color{red}{\bf \frac{-1}{u^{2}}}\cdot \color{blue}{\bf \frac{1}{2\sqrt x}} = \dfrac{-1}{\color{green}{\bf(1+\sqrt x)}^2}\cdot \dfrac{1}{2\sqrt x} = \dfrac{-1}{2\sqrt x(1+\sqrt x)^2}$$
