The concave mirror formula is

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ .... (i)

Since, for a given concave mirror, focal length $f$ is fixed, we partially differentiate Eq. (i) to get

$-\frac{\delta v}{v^2}=-\frac{\delta u}{u^2}=0$ or $\delta v=-{\left(\frac{v}{u}\right)}^2 \delta u$ ...... (ii)

Multiplying Equation. (i) by $u$ , we get

$\frac{u}{v}+1=\frac{u}{f}$ or $\frac{v}{u}=\frac{f}{u-f}$ ..... (iii)

Using Equation. (iii) in Equation. (ii), we get

$\delta v=-{\left(\frac{f}{u-f}\right)}^2\delta u$

Given $\delta u=b.$ Therefore

$\delta v=-{\left(\frac{f}{u-f}\right)}^{2}b$

The negative sign shown that image is longitudinally inverted. The magnitude of the size of the image is

$\left|\delta v\right|=b{\left(\frac{f}{u-f}\right)}^2$