Chain rule clarification? I am not understanding how if you have some function that is f(g(x)) that the derivative of that function is f '(g(x)) g '(x). This doesn't make sense to me because shouldn't f'(g(x)) be f'(g(x))? I know this must be an incredibly stupid question but can someone please help? For instance, sqrt((3x+5)^2). The rule says to differentiate ((3x+5)^2) which is 2(3x+5) and then multiply it by 3, or the derivative of 3x+5, to get 6(3x+5). Now my question is pretty much this, if the first step of differentiating this is to differentiate the entire thing, then aren't we done because the derivative of ((3x+5)^2) is 2(3x+5)? 
 A: $2(3x+5)$ is not just the derivative of $(3x+5)^2$; it is derivative of $(3x+5)^2$ with respect to $3x+5$.  Which is not equal to the derivative with respect to $x$ , that being: $6(3x+5)$
When asked for a derivative it is always with respect to some variable parameter; either explicitly or implied by context of the discussion.  
Thus when asked to find the derivative of $f(g(x))$ it is implied to be with respect to $x$, the independent variable of the discussion.

The following are equivalent expressions
$$\begin{align}
 \frac{\operatorname d f(g(x))}{\operatorname d x}  & =\frac{\operatorname d f(g(x))}{\operatorname d g(x)}\cdot\frac{\operatorname d g(x)}{\operatorname d x} 
\\[2ex]
 [f(g(x))]' & =f'(g(x))\cdot g'(x)
\\[2ex]
 [f\circ g]'(x) & = [f'\circ g \cdot g'](x)
\\ & = [f'\circ g](x)\cdot g'(x)
\end{align}$$
The main point is that $f'(g(x))$ is the derivative of $f(g(x))$ w.r.t. $g(x)$; and not the derivative of $f(g(x))$ w.r.t. $x$.
$$f'(g(x)) = \frac{\operatorname d f(g(x))}{\operatorname d g(x)}{\Large \neq} \frac{\operatorname d f(g(x))}{\operatorname d x}$$
A: So we have the function:
$$f(x) =(3x+5)^2$$
Let $g(x) = 3x + 5$, therefore we have that $g'(x) = 3$
We have that:
$$f(x) = (g(x))^2$$
$$f'(x) = 2*(g(x))*g'(x)$$
$$f'(x) = 2(3x+5)*3 = 6(3x+5)$$
Hope this helped.
A: A helpful way of thinking about differentiating chain functions is to think about them like they're your grandma wrapped in a big winter coat.
In this case your grandma is :
$$g(x)= 3x+5$$
Her big poofy coat is $$f(x) = (grandma)^2$$
(doesn't she look nice and cozy?)
...or in more formal terms $$f(x) = (g(x))^2$$
Keep in mind that f isn't just your grandma and isn't just her coat, it's your grandma in her coat.
So when someone asks you to differentiate your grandma in her coat, don't just differentiate her coat (as you would be doing in your last question). Go ahead and differentiate the coat, followed by your grandma. Then you've differentiated the entire function and you're all set.
So,
$$f'(x) = 2(3x+5)*3$$
$$f'(x) = 6(3x+5)$$
