How to rewrite $\frac{d}{d(x+c)}$? I would like to know how to rewrite the following equations:
$$
\frac{d (f(x))}{d(x+c)} =0\\
\frac{d^2 (f(x))}{d(x+c)^2} =0\\
$$
Here $x$ is a variable, $c$ is a constant and $f(x)$ is a function of x.
I would also like to know the reasoning behind the answer.
 A: Use the chain rule. Define $u = x + c$ then use the fact that $$\frac{d\cdot}{dx} = \frac{du}{dx} \frac{d\cdot}{du}$$ where the $\cdot$ represents any function, so
$$\frac{df}{dx} = \frac{du}{dx} \frac{df}{du}$$
It also follows that 
$$
\begin{array}{rcll}
\frac{d^2f}{dx^2} &=& \frac{d}{dx} (\frac{df}{dx}) &\quad\mbox{definition of 2nd derivative} \\
&=& \frac{d}{dx} \big(\frac{du}{dx} \frac{df}{du}\big) &\quad\mbox{using the result above} \\
&=& \frac{d}{dx} \big(\frac{du}{dx} \big)\frac{df}{du} + \frac{du}{dx} \frac{d}{dx} \big(\frac{df}{du}\big) &\quad\mbox{using the rule for the derivative of a product} \\
&=& \frac{d^2u}{dx^2} \frac{df}{du} + \frac{du}{dx} \frac{du}{dx} \frac{d}{du} \big(\frac{df}{du}\big) &\quad\mbox{using various results from above} \\
&=& \frac{d^2u}{dx^2} \frac{df}{du} + (\frac{du}{dx})^2 \frac{d^2f}{du^2} &\quad\mbox{simplifying}
\end{array}
$$
So, in summary,
$$\frac{df}{dx} = \frac{du}{dx} \frac{df}{du}$$
and
$$\frac{d^2f}{dx^2} = \frac{d^2u}{dx^2} \frac{df}{du} + (\frac{du}{dx})^2 \frac{d^2f}{du^2}$$
The above is correct and valid for any $u(x)$ but it's written in a somewhat backwards way. You already have $\frac{df}{dx}$ and $\frac{d^2f}{dx^2}$ and want to find $\frac{df}{du}$ and $\frac{d^2f}{du^2}$. Well, that's simple algebra now to get those from the above. It's even simpler with the specific example of $u = x + c$ because, then, $\frac{du}{dx} = 1$ and $\frac{d^2u}{dx^2} = 0$, so
$$\frac{df}{du} = \frac{df}{dx}$$
and
$$\frac{d^2f}{du^2} = \frac{d^2f}{dx^2}$$
for that particular example.
A: Introducing $y=x+c$ one gets
$$
0 
= \frac{d(f(x))}{d(x+c)}
= \frac{d(f(y-c))}{dy}
= f'(y-c) \cdot 1
= f'(x)
= \frac{d(f(x))}{dx}
$$
For the second derivative we get:
$$
0 
= \frac{d^2(f(x))}{d(x+c)^2}
= \frac{d}{d(x+c)} \frac{d(f(x))}{d(x+c)}
= \frac{d(f'(x))}{d(x+c)} = (f')'(x) = f''(x)
= \frac{d^2(f(x))}{dx^2}
$$
A: $$\frac { d }{ dx } \left( f\left( x \right) \left( x+c \right) ^{ -1 } \right) =0\\ \frac { df\left( x \right)  }{ dx } \left( x+c \right) ^{ -1 }+f\left( x \right) \frac { d\left( \left( x+c \right) ^{ -1 } \right)  }{ dx } =0\\ \frac { df\left( x \right)  }{ dx } \frac { 1 }{ \left( x+c \right)  } -f\left( x \right) \frac { 1 }{ { \left( x+c \right)  }^{ 2 } } =0$$
