# What is a closed form expression for the ∂/∂w(∂t/∂w) if w(t) is complicated function?

Lets say we have a trigonometric function w(t) that can not be inverted as t(w).

The derivative ∂t/∂w can be calculated as 1/(∂w(t)/∂t). What is a closed form expression for the second derivative ∂/∂w(∂t/∂w) ?

Numerically do you start from the closed form or there is easier way ?

This represents the group delay dispersion and third order dispersion.

If I understand correctly you are interested in a closed form for the second derivative of the inverse function expressed in terms of other information. Then:

$f(f^{-1}(x))=x$

$f'(f^{-1}(x))\cdot(f^{-1})'(x)=1$ (differentiating the previous identity)

$f''(f^{-1}(x))\cdot((f^{-1})'(x))^{2}+f'(f^{-1}(x))(f^{-1})''(x)=0$ (differentiating again)

so $(f^{-1})''(x)=\frac{-f''(f^{-1}(x))\cdot((f^{-1})'(x))^{2}}{f'(f^{-1}(x))}=\frac{-f''(f^{-1}(x))}{(f'(f^{-1}(x)))^{3}}$

Note that you need not invert the function to compute this. If $f(a)=b$ and you want the derivative of $f^{-1}$ at $x=a$ then $f^{-1}(a)=b$ so the inverse drops from the closed form.

• Yes, thats it. thank you. So ∂/∂w(∂t/∂w) = - ((∂^2w(t)/∂t^2))/((∂w(t)/∂t))^3 Commented Jul 10, 2014 at 8:04
• You're welcome. Yes that is the formula. Make sure you are careful about the point of evaluation in the derivatives. Commented Jul 10, 2014 at 8:13