Two variables u and v are connected by the equation 1/u + 1/v = 1/f where f is a constant. Given that f=10cm

this is the thin lens equation where u and v are the object and image distances and f is the focal length

if f is positive you have a converging mirror; f is negative for diverging mirrors

note added later:

take the time derivative of each term:

-1/u^2 du/dt - 1/v^2 dv/dt = -1/f^2 df/dt

since f is a constant, df/dt =0 and then -1/u^2 du/dt - 1/v^2 dv/dt =0

when u=40, we find the value of v:

1/40 + 1/v = 1/10

1/v = 1/10 - 1/40
1/v = 3/40
v= 40/3 = 13.3

if du/dt = +2, then

-1/u^2 du/dt - 1/v^2 dv/dt =0

-1/40^2 x 2 - 1/13.3^2 dv/dt =0

dv/dt = -13.3^2/40^2 du/dt = -13.3^2/40^2 x 2 = -0.22cm/sec

this means that if the object is moving away from the lens at the rate of 2cm/s, when the object is 40m away, the image is getting closer to the lens at the rate of 0.22cm/s

Its all rite. this is a simple problem.
First using the given equation and the values u=40 and f=10 , find v , which turns out to be 40/3.

next differentiate the given eqn with respect to t

you'll have the following eqn -1/u^2 x du/dt -1/v^2 x dv/dt = 0
simplifying it , you get
dv/dt = -v^2/u^2 x du/dt
now u=40 and v=40/3 and it is given that , du/dt is -2cm/sec
substitute it here
you get dv/dt something like 0.22 cm/sec. i hope i've helped.
im looking forward for those 10 pts.