Let's consider a circle whose area is increasing at a constant rate of 5m²/s.We are to find the rate of change of radius of the circle at the instant when the radius of the circle is 2.5cm.
This is how I proceeded.
*Let area of the circle at time t be A m² and it's radius be r m.
A=πr²
or,dA/dt=2πrdr/dt
Now,this is where my problem arises.
We replace dA/dt by 5 and r by 2.5.
My question is that why don't we replace the term 'dr/dt' by 'd(2.5)/dt'…? *
To understand the correct substitutions in related rates problems, let's look at what the differential equation you arrived at actually tells us. Your implicit differentiation was correct which gave us: $$\frac{dA}{dt}=2\pi r\frac{dr}{dt}$$ We can then rewrite this expression to get an explicit function for $\frac{dr}{dt}$: $$\frac{dr}{dt}=\frac{1}{2\pi r}\left(\frac{dA}{dt}\right)$$ Knowing $\frac{dA}{dt}=5$ for all $r$, we have: $$\frac{dr}{dt}=\frac{5}{2\pi r}$$This shows that $\frac{dr}{dt}$ is just a function of $r$, meaning that when we plug in $r=2.5$ to your original equation we are evaluating the function for $\frac{dr}{dt}$ at $r=2.5$ which can be written as $\frac{d(r=2.5)}{dt}$. However, in your solution substituting $\frac{d(2.5)}{dt}$ for $\frac{dr}{dt}$ is redundant because $\frac{dr}{dt}$ at $r=2.5$ is what we are actually solving for. It would be similarly redundant if we were evaluating $y$ for an algebraic equation $y=f(x)$ at a value $x=c$, to give the solution as $\operatorname{y}(c)=f(c)$ instead of $y=f(c)$.
I think your problem is in the notation conventions.
If you write everything in terms of functions the problem disappears. Suppose the area at time $t$ is $A(t)$ and the radius at time $t$ is $r(t)$. Those two functions are related: $$ A(t) = \pi r(t)^2. $$ Differentiating with respect to $t$, using the chain rule on the right: $$ A'(t) = 2 \pi r(t) r'(t) . $$ Then the substitutions you make are straightforward and routine: $$ 5 = 2 \pi \times 5 \times r'(t) . $$ Then solve that for $r'(t)$. (You don't know and can't find out what time that actually happens, but you don't care.)
If you want to use the differential notation (which is valuable and worth knowing how to do correctly) you should think of "$dr$" as "a little bit of $r$", not as "$d$ times the value of $r$".
Then the equation $$ (dA) = 2 \pi r (dr) $$ tells you the relationship between a small change in the radius and a small change in the area. That makes geometric sense if you think of surrounding a circle of radius $r$ by an annulus of width $dr$ and seeing how the area changes.
Dividing by a small change $dt$ in $t$ turns those small changes into instantaneous rates of change.
