# How to rewrite $\frac{d}{d(x+c)}$? [closed]

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.

• There is no equality sign so they are not equations anyway. Jul 28, 2015 at 10:14
• why is this getting close votes? If the OP knew how to write a perfect expository paper on Leibniz notation, they presumably wouldn't need to ask the question. Jul 28, 2015 at 10:32
• I'm still trying to figure out the answers. When I find one to be the most useful answer to my question, I will mark this question as answered. Jul 28, 2015 at 10:45
• Use $d(x+c)=dx$ and you are done.
– user65203
Jul 28, 2015 at 13:39

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.

• The problem here is that it is not clear from notation if it is $(x+c)^2$ or the second derivative with respect to $(x+c)$. Jul 28, 2015 at 10:22
• @mathreadler If it were the second derivative one usually writes $$\frac{d^{\mathbf{2}}f(x)}{dx^2}$$ though. But you may be correct still.
– Eff
Jul 28, 2015 at 10:23
• Yes Eff, in the case you talk about the derivative is with respect to x, and there can be no confusion from notation. But what was intended in the question was the various derivatives with respect to $(x+c)$ (in some sense). Jul 28, 2015 at 10:38
• @mathreadler If that's the case, then there's a missing superscript 2 in the "numerator", i.e., it should be $\frac{d^2}{d(x+c)^2}$ rather than $\frac{d}{d(x+c)^2}$. Jul 28, 2015 at 10:52
• Yes wltrup you are right, but there were several other things missing in the original formulation also. So the question is, what was intended? It could be many things. Jul 28, 2015 at 10:54

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

• I think in the second case OP is meaning the second derivative (see d^2 on the top) Jul 28, 2015 at 14:10
• Seems like an update. Thanks.
– mvw
Jul 28, 2015 at 14:11

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