How to find derivative of $\frac{1}{x^2 + x}$ and how to understand the chain rule? I don't know how to do this one. I failed my pre-calc/calc course and I need someone to explain how to do this. The materials I have are not helping me very much.
If someone can explain the chain rule to me as well that would be great.
$$\frac{1}{x^2 + x}$$
 A: $f(x)=\frac1x,g(x)=x^2+x\Rightarrow f\circ g(x)=\frac{1}{x^2+x}$
Now $(f\circ g)'(x)=f'(g(x))g'(x)=-\frac{1}{(x^2+x)^2}(2x+1)=-\frac{2x+1}{(x^2+x)^2}$
A: You can write $\frac{1}{x^2+x}$ as $(x^2+x)^{-1}$.
The chain rule is $f(g(x))'=f'(g(x))g'(x)$. You can think of a composite function as being built up from an "inside" function and an "outside" function.
The "inside" function is $g(x)$ in the chain rule. In the example it would be $x^2+x$. If we let $y=g(x)=x^2+x$, we can think of the "outside" function as $f(y)=y^{-1}$.
Now to use the chain rule, first we have the $f'(g(x))$ part. I often describe this as "take the derivative of the outside function but leave the inside the same.  In the example,
$$f'(g(x))=-1(x^2+x)^{-2}.$$
The next part is $g'(x)$ which is the derivative of the inside function. So for the example
$$g'(x)=2x+1.$$
So altogether the chain rule says " take the derivative of the outside function while leaving the inside the same, the multiply by the derivative of the inside".
So, put this together gives
$$\begin{align}
f(g(x))' & =f'(g(x))g'(x)\\
& =-1(x^2+x)^{-2}(2x+1)\\
& =-\frac{2x+1}{(x^2+x)^2}.
\end{align}$$
A: Whenever you take the derivative of "a bunch of stuff under $1$" rewrite it so you can use the power rule. In this case:
$$(x^2+x)^{-1}$$
Recall the power rule subtracts $1$ from the exponent (in this case $-1$) and then places the orignal in front of the function. For example:
$$\frac{d}{dx}u^{-1}=-u^{-2}$$
Now in your case you have "a bunch of stuff" where we had $x$ in the previous example. The chain rule allows us to deal with that stuff separately. First set your stuff to equal $u$:
$$u=x^2+x$$
Then plug $u$ back into the original equation:
$$u^{-1}$$
Evaluate $u$ using the power rule as we did above:
$$-u^{-2}$$
The final step when using the chain rule is to take the derivative of the stuff we set $u$ to equal and multiply it by the derivative of $u$:
$$\frac{d}{dx}(x^2+x)=2x+1$$
$$-u^{-2}(2x+1)$$
Now just plug our stuff back into $u$:
$$-(x^2+x)^{-2}(2x+1)$$
You can then rewrite this into a more readable form. First note that we have a negative exponent. This just means "the stuff in the exponent under $1$" to whatever power. This is essentially the opposite of the first move we performed to use the power rule in the first place. So:
$$-(x^2+x)^{-2}=-\frac{1}{(x^2+x)^2}$$
All we have left to do is multiply:
$$-\frac{1}{(x^2+x)^2}*2x+1=\frac{2x+1}{(x^2+x)^2}$$
A: you can find the derivative of ${1 \over x^2 + x}$ by splitting it to 
$${1 \over x^2 + x} = {1 \over x}- {1 \over x + 1}$$ you still have to use chain rule in finding the derivative of $1 \over x + 1$ though.
in the case we have here, which is the graph of $ y = {1  \over x + 1}$ is a translation of $y = {1 \over x},$ the derivatives are also translated. that is 
$({1 \over x})^\prime = -{1 \over x^2 }$ implies 
$({1 \over x + 1})^\prime = -{1 \over (x + 1)^2}$. an application of chain rule is hidden there but in a simpler form of translation. 
A: First note that

$$\frac{1}{x^2 + x}= \left(x^2+x\right)^{-1}$$

Also note that the chain rule is defined as

$$ \frac{d}{dx}f(g(x))=\frac{d}{dg(x)}f(g(x))\frac{d}{dx}g(x)$$

Here are some other derivative rules that I used

$$ \frac{d}{dx}\left[x^{n}\right]=nx^{n-1} $$
  $$ \frac{d}{dx}\left[f(x)\pm g(x)\right]=\frac{d}{dx}f(x)\pm\frac{d}{dx}g(x) $$

So now we have

$$ \frac{d}{dx}\left[\left(x^2+x\right)^{-1}\right]=-\left(x^2+x\right)^{-2}\frac{d}{dx}\left[x^2+x\right]$$
  $$ =-\left(x^2+x\right)^{-2}\left(\frac{d}{dx}\left[x^2\right]+\frac{d}{dx}[x]\right) $$
  $$ =-\left(x^2+x\right)^{-2}\left(2x+1\right) = -\frac{2x+1}{\left(x^2+x\right)^2}$$

