For a quantity V depending on a changing depth h(t), the chain rule gives dV/dt=V′(h) dh/dt. A draining tank has dh/dt<0, so signed dV/dt is negative; an outflow magnitude is its negative. Draw the geometry and name the instantaneous depth before substituting numbers. Do not treat the rate as a static volume divided by elapsed time unless the situation actually specifies a constant average rate.
A spherical tank of radius R has cross-section radius squared R²−(R−h)²=2Rh−h² at depth h from the bottom. Integrating these circular areas gives the cap volume V(h)=π(Rh²−h³/3), for 0≤h≤2R. Its derivative π(2Rh−h²) is the current cross-sectional area. At R=3,h=1,dh/dt=−1/4, signed volume rate is −5π/4 and outflow magnitude 5π/4. The formula works for both shallow and deep caps within the stated range.
If f and g are continuously differentiable near zero, f(0)=g(0)=0 and g′(0)≠0, then f(x)/g(x) tends to f′(0)/g′(0). This follows from f(x)=f′(0)x+o(x) and g(x)=g′(0)x+o(x); continuity of g′ keeps the quotient defined nearby except at zero. Filling in this limit gives a continuous extension. The argument does not require f′(0) nonzero, and it does not prove the extended quotient differentiable.
For f(x)=x|x| and g(x)=x, both are continuously differentiable, f(0)=g(0)=0 and g′(0)=1. Their quotient for x≠0 is |x|, which extends continuously at zero but has unequal one-sided derivatives there. For (f²−f)/(2g−g³), factor to (f/g)(f−1)/(2−g²); its removable limit is −f′(0)/(2g′(0)). Additional factors approach −1 and 2. Distinguish the limit of the quotient from the derivative of its extension.
