CHAIN RULE IN LEIBNIZ NOTATION
If f and g are differentiable functions, then the chain rule explains how to differentiate the composite g o f. Now let us give separate names to the dependent and independent variables of both f and g so that we can express the chain rule in the Leibniz notation. Suppose that
y = g(u) and u = f(x).
In this way we see that y is a function of u and that u in turn is a function of x. Putting these two processes together produces the composite function
g o f that determines y as a function of x: y=g(f(x)).
In particular, y is a function of u and also of x; the intermediate variable u is a function of x. We may therefore discuss the rates of change of y with respect to both u and x, as well as the rate of change of u with respect to x.
The chain rule is an algebraic relation between these three rates of change. In fact, the chain rule says that the first rate of change is the product of the other two.
Translating the chain rule into Leibniz notation.
Here is the chain rule again, still in the prime notation of Lagrange.
Chain Rule in Prime Notation

( 
g(f(x))

) 
= 
g'(f(x))f '(x)


The function u = f(x) expresses the dependence of u on x; the rate of change of u with respect to x is the derivative.
The function y = g(u) expresses the dependence of y on u; the rate of change of y with respect to u is the derivative.
The composite function y = g(f(x)) expresses the dependence of y on x; the rate of change of y with respect to x is the derivative.
Substituting these translations into the chain rule, we obtain the following formula.
Chain Rule in Leibniz Notation

The chain rule is easy to remember in this setting...it appears that we can just cancel out the du's!!!! (Of course, this really makes no literal sense, since du itself has no explicit numerical significance. Some people like to attach significance to the separate symbols dx, du, and dy by studying things called differentials and infinitesimals. We will not discuss these topics in these terms.)
V = 4r^{3}/3

The Role of Mulitplication in the Chain Rule. Whether you prefer prime or Leibniz notation, it's clear that the main algebraic operation in the chain rule is multiplication. In order to illustrate why this is true, think about the inflating sphere again.
As air is pumped into the balloon, the volume and the radius increase. Both volume and radius are functions of time. The volume formula for a sphere also expresses the volume as a function of radius: V = 4r^{3}/3. Given a time instant t, a radius r(t) is determined (namely, the radius at time t). The radius value in turn determines the volume. In this way, the dependence of volume on time is expressed as a composite function.
The chain rule now relates the rates of change of volume with respect to radius and time to the rate of change of radius with respect to time.

= 
( 

)( 

) 
= 
( 
4r^{2}

)( 

) 
Sample Problem. Suppose that we want to inflate a spherical balloon in such a way that the radius increases at a constant rate of 2 inches per second. At what rate is should air be pumped into the balloon when the radius is 1 foot? When the radius is 20 feet?
Solution. The rate at which air should be pumped into the balloon is the rate of change of the volume of the balloon with respect to time. This quantity is represented symbolically by the derivative dV/dt. We measure this rate of change in cubic feet per second. A rate of change of 2 inches per second for the radius translates into 1/6 foot per second. This means that dr/dt is to be held constant at 1 foot for each 6 second time interval. When the radius r is 1 foot, we find the necessary rate of change of volume using the chain rule relation as follows.

= 
( 
4r^{2}

)( 

) 

= 
(4(1 foot)^{2})(1 foot/6 seconds)


= 
(2/3) ft^{3}/sec 2.094 cubic feet per second

When the radius r is equal to 20 feet, the calculation proceeds in the same way.

= 
( 
4r^{2}

)( 

) 

= 
(4(20 feet)^{2})(1 foot/6 seconds)


= 
(800/3) ft^{3}/sec 837.758 cubic feet per second

As we saw before, the rate of change of volume depends heavily on the radius at the instant under consideration, even if the rate of change of radius is fixed!!
Answer. In order to inflate a spherical balloon in such a way that the radius is increases at a constant rate of 2 inches per second, then when the radius is one foot, the air must be pumped into the balloon at a rate of approximately 2.094 cubic feet per second. (We could probably do this with an simple hand pump.) When the radius is 20 feet, in order to maintain a rate of increase of two inches per second in the radius, air must be pumped into the balloon at a rate of approximately 837.758 cubic feet per second. (Imagine trying to push that much air with a hand pump!!)
Sample Problem. Here is a more realistic problem. Suppose that air is being pumped into a spherical balloon at a constant rate of 4 cubic feet per second. At what rate is the radius increasing when the radius is 1 foot? When the radius is 20 feet?
Solution. The fact that air is being pumped into the balloon at four cubic feet per second is expressed symbolically in this way.
We are asked to determine the rate of change of radius with respect to time. Thus we should compute dr/dt. The answer depends on the particular radius at the instant in question. We use the relation that was derived from the chain rule in the previous example.

= 
( 
4r^{2}

)( 

) 
This time the approach is to substitute in the given values of dV/dt and r and then solve for dr/dt. When r=1 foot, the calculation goes this way.
4 ft^{3}/sec

= 
( 
4(1 foot)^{2}

)( 

) 
Solving for dr/dt, we compute as follows.

= 
4 ft^{3}/sec 

4 ft^{2} 

= 
(1/) ft/sec 0.3183 ft/sec 
We repeat the calculation for r=20 feet.
4 ft^{3}/sec

= 
( 
4(20 feet)^{2}

)( 

) 
Solving for dr/dt, we compute as follows.

= 
4 ft^{3}/sec 

4(400) ft^{2} 

= 
(1/400) ft/sec 0.0008 ft/sec 
Answer. If air is being pumped into a spherical balloon at a constant rate of 4 cubic feet per second and the radius is one foot, then the radius is increasing at a rate of approximately 0.318 feet per second, or about 3.820 inches per second. When the radius is 20 feet, the radius is increasing at approximately 0.0008 ft/sec, or just 0.0095 inches per second. This is approximately equal to a rate of increase of 34.38 inches per hour.
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William A. Bogley
Robby Robson