# How to represent this symbol - the vertical line denoting derivative at a point?

Below is Taylor series defined that I'm attempting to represent as MathJax :

Closest I have is :

$\sum \limits_{k=0}^\infty \frac {f^k(0)} {k!} x^k = f(0) + \frac {df} {dk}x_0+ \frac 1 {2!} \frac {{d^2}f} {d{x^2}}x_0^2$

How to represent the long vertical line after each $\frac {df} {dx}$ that is before $x$ ?

• Try {a\over b}\Big| $${a\over b}\Big|$$ – Gerry Myerson May 12 '17 at 7:19
• @GerryMyerson "{a\over b}\Big|_{0}" $${a\over b}\Big |_{0}$$ – mrnovice May 12 '17 at 13:28

Both $$f(0) + \left.\frac {df}{dx}\right|_0 x+ \frac 1{2!} \left. \frac {{d^2}f} {d{x^2}}\right|_0x^2 + \dots$$ and $$f(0) + \left.\frac {df}{dx}\right\rvert_0 x+ \frac 1{2!} \left. \frac {{d^2}f} {d{x^2}}\right\rvert_0x^2 + \dots$$ seem to me like reasonable approximation.
$$f(0) + \left.\frac {df}{dx}\right|_0 x+ \frac 1{2!} \left. \frac {{d^2}f} {d{x^2}}\right|_0x^2 + \dots$$
$$f(0) + \left.\frac {df}{dx}\right\rvert_0 x+ \frac 1{2!} \left. \frac {{d^2}f} {d{x^2}}\right\rvert_0x^2 + \dots$$. I simply copied part of your text and added \left. and \right| or \right\rvert.