Derivation in bohm's Quantum Theory

1. Aug 17, 2012

A_B

halfway page 41 Bohm obtains for the action variable
$$J = 2\int_{a(E)}^{b(E)} dq \sqrt{2m[E-V(q)]}$$
Then he obtians the partial derivative to E "by a well-known theorem of the calculus":

$$\frac{\partial J}{\partial E} = 2\left\{ \sqrt{2m\left[E-V(q)\right]} \right\}_{q=b} \frac{\partial b}{\partial E} - 2\left\{ \sqrt{2m[E-V(q)]} \right\}_{q=a} \frac{\partial a}{\partial E} + 2 \int_a^b \sqrt{\frac{m}{2[E-V(q)]}} dq.$$

What is this "well-known theorem"?

Thanks,
A_B

2. Aug 17, 2012

Jorriss

3. Aug 17, 2012

A_B

thanks! ...Never came across that one before...

A_B