How to rewrite $\frac{d}{d(x+c)}$?

Question

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.

Answer 1

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.

Answer 2

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: $$ 0 = \frac{d^2(f(x))}{d(x+c)^2} = \frac{d}{d(x+c)} \frac{d(f(x))}{d(x+c)} = \frac{d(f'(x))}{d(x+c)} = (f')'(x) = f''(x) = \frac{d^2(f(x))}{dx^2} $$

Answer 3

$$\frac { d }{ dx } \left( f\left( x \right) \left( x+c \right) ^{ -1 } \right) =0\\ \frac { df\left( x \right) }{ dx } \left( x+c \right) ^{ -1 }+f\left( x \right) \frac { d\left( \left( x+c \right) ^{ -1 } \right) }{ dx } =0\\ \frac { df\left( x \right) }{ dx } \frac { 1 }{ \left( x+c \right) } -f\left( x \right) \frac { 1 }{ { \left( x+c \right) }^{ 2 } } =0$$