Longitudinal magnification I want to prove that if an object is small in length and lying along the principal axis then
$$M = -\frac{dv}{du} = -\left(\frac{v}{u}\right)^2$$
Where, $M$ is the longitudinal magnification.

 A: In geometrical optics the following relation between the longitudinal positions of object and image (respectively $u$ and $v$) together with the focal length $f$ is valid:
$$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$$
If the object is small and it has one of its ends at $u_1$, with the corresponding image at $v_1$, we can calculate the position of the image of the other end, $v_2$, in an approximate way using derivatives:
$$v_2 \approx v_1 + \frac{dv}{du} (u_2 - u_1)$$.
where $\frac{dv}{du}$ is the derivative calculated at $u_1$.
The longitudinal magnification is the ratio between the length of the image and the length of the object:
$$ M = \left|\frac{v_2 -  v_1}{u_2 -  u_1}\right|, $$ 
and using the approximate equation for the position in terms of the derivatives that we wrote above we see that
$$M \approx \left| \frac{dv}{du} \right|.$$
An easy way to calculate the derivative is considering that the variation of the quantity $\frac{1}{u} + \frac{1}{v}$ is zero for any variation of the position of the object $u$ and the corresponding variation of the position of the image $v$. So:
$$d \left (\frac{1}{u} + \frac{1}{v}\right) = 0$$.
We can express the variation using the variations of $u$ ($du$) and $v$ ($dv$) as
$$d \left (\frac{1}{u} + \frac{1}{v}\right) = -\frac{1}{u^2}\,du -\frac{1}{v^2}\,dv$$
still equal to zero. From $-\frac{1}{u^2}\,du -\frac{1}{v^2}\,dv = 0$ we obtain the expression for $\frac{dv}{du}$:
$$\frac{dv}{du} = - \frac{v^2}{u^2}$$
The longitudinal magnification is then 
$$M_{\mathrm{long.}} = \frac{v^2}{u^2}$$
Since the transverse magnification is 
$$M_{\mathrm{transv.}} = \frac{v}{u}$$
then
$$M_{\mathrm{long.}} = M_{\mathrm{transv.}}^2$$
