This is a very elementary question, but it has caused be a lot of confusion these last days when working on some problems.

If I write:

$f'(-x)$, does it mean that I have the function f, then i differentiate it with respect to x, and then I insert -x for x, or does it mean that I first insert (-x) in the function and then differentiate with respect to x?

The same question goes for:


Then what does $\frac{\partial u(-x,y)}{\partial x}$mean?

I can either first differentiate with respext to x to get: $2*x*y$ and then insert -x. to get $-2*x*y$. Or I can insert the -x first, and then differentiate so I get $2*x*y$. What is correct?

Also, if one of them is correct, what is the correct notation for the other kind?

  • $\begingroup$ $f'(-x)$ is probably equal to $\frac{df}{dx}\large|_{x=-x}$. You differentiate first and then insert $-x$ for $x$. Same thing for the partial. $\endgroup$ – Shahar Sep 9 '14 at 23:01
  • $\begingroup$ @Shahar No, it's a different meaning for the partial. The notations do not have the same structure. $\endgroup$ – Graham Kemp Sep 9 '14 at 23:36

$\begin{align} \mathrm{Since} \\[2ex]f'(g(x)) & = \dfrac{\operatorname{d}f(g(x))}{\operatorname{d}g(x)} \\[1ex] & = \left.\dfrac{\operatorname{d}f(z)}{\operatorname{d}z}\right|_{z=g(x)} \\[1ex] & = \left.f'(z)\right|_{z=g(x)} \\[2ex] \mathrm{Then} \\[2ex]f'(-x) & =\dfrac{\operatorname{d}f(-x)}{\operatorname{d}(-x)} \\[1ex] & = \left.\dfrac{\operatorname{d}f(z)}{\operatorname{d}z}\right|_{z=-x} \\[1ex] & = \left.f'(z)\right|_{z=-x} \\[2ex] \mathrm{Similarly} \\[2ex]u^{(1,0)}(-x,y) & =\dfrac{\partial u(-x,y)}{\partial(-x)} \\[1ex] & = \left.\dfrac{\partial u(z,y)}{\partial z}\right|_{z=-x} \\[1ex] & = \left.u^{(1,0)}(z,y)\right|_{z=-x} \\[1ex] & = \left.2zy\right|_{z=-x} \\[1ex] & = -2xy \\[2ex] \mathrm{HOWEVER} \\[2ex] \dfrac{\partial u(-x, y)}{\partial x} & = -\dfrac{\operatorname{d}(-x)}{\operatorname{d} x}\cdot \dfrac{\partial u(-x,y)}{\partial (-x)} \\[1ex] & = - \dfrac{\partial u(-x,y)}{\partial (-x)} \\[1ex] & = 2xy \\[5ex] \dfrac{\partial (-x)^2y}{\partial x} & = 2xy \end{align}$

  • $\begingroup$ Is this correct notation?: $\dfrac{\partial u(-x,y)}{\partial x}$, here insert -x and then differentiate. And $\dfrac{\partial u}{\partial x}(-x,y)$, here we first differentiate and then insert? $\endgroup$ – user119615 Sep 10 '14 at 13:07

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