Suppose we are heating a spring whose form is the curve w = f (x, y, z). Suppose the temperature at a point t is given by x = g (t), y = h (t), z = k (t) where t is time. Therefore, we could consider w to be a function of t and w = f (g (t), h (t), k (t)). If we needed to, we could find the rate at which f changes with respect to t by differentiating f with respect to t. This is perfectly fine, if substituting g, h, and k into f is fairly easy. If it is not, then we have to use the chain rule.

THEOREM:

t is the independent variable.

Sometimes it is useful to draw a tree diagram of how the chain rule will be set up.

Since w is a function of two variables, then when we look for the derivatives, we will be finding partial derivatives. On a tree diagram, drawing two branches coming out of the w node will indicate this fact.

Since x and y are functions of one variable, then we will be finding the ordinary derivative. On the tree diagram, drawing one branch coming out or the x and y node will indicate this fact.

To find dw/dt, sum up the derivatives of each branch. (See figure 1)

EXAMPLE 1:

Find dw/dt when w = x ² + y ² and x = cos t + sin t and y = t ³ - t ⁴.

SOLUTION:

This derivative is not complete. For it to be complete, it has to be in terms of t alone. No x's or y's left in it. So substitute in for x and y.

EXAMPLE 2:

Find dw/dt when w = ln (x ² + y ² + z ²) and x = cos t, y = sin t, and z = 4t².

SOLUTION:

Now w has 3 variables, so the tree diagram will look like the following figure (See figure 2) and the chain rule will be the following.

figure 2

Suppose w = f (x, y, z) represents the temperature at a point (x, y, z) on a sphere, then x, y, and z are functions of 2 variables, r and s which could represent the longitude and latitude. If x = g (r, s), y = h (r, s), and z = k (r, s), then w = f (g (r, s), h (r, s), k (r, s)) and the derivatives would be the following.

Here are the tree diagrams for these partial derivatives.

DEFINITION:

If w = f (x, y), x = g (r, s) and y = h (r, s), then

DEFINITION:

If w = f (x) and x = g (r, s), then

and the tree diagram is the following.

EXAMPLE 3:

Find ¶ z / ¶ r and ¶ z / ¶ q for z = 4e ^x ln y, x = ln (r cos q ) and y = r sin q .

SOLUTION:

=(4cos q )(ln (r sin q )) + 4cos q

EXAMPLE 4:

Draw the tree diagram and write the chain rule for dz/dt for z = f (u, v, w), u = g (t), v = h (t), and w = k (t).

SOLUTION:

z is a function of 3 variables- u, v, w, so each variable will be a branch. u, v, and w are functions of a single variable, so u, v, and w will have a single branch coming out of the node. Here the tree diagram for the derivative.

and the chain rule formula is the following.

EXAMPLE 5:

Draw the tree diagram and write the chain rules for ¶ z/¶ t and ¶ z/¶ s for z = f (x, y), x = g (t, s), and y = h (t, s).

SOLUTION:

z is a function of 2 variables, so there will be two branches coming out from z. x and y are both functions of two variables, so there will be two branches coming out of x and y. So here is the tree diagram for these partial derivatives.

Here are the chain rules for these two derivatives.

EXAMPLE 6:

Draw the tree diagram and write the chain rule for ¶ w/¶ p for w = f (x, y, z, v), x = g (p, q), y = h (p, q), z = j (p, q), and v = k (p, q).

SOLUTION:

Since w is a function of four variables, the tree will have four branches coming out of the w node. Since each node- x, y, z, v - is a function of two variables, so each one of those nodes will have two branches coming out of them. Here is the tree diagram for this partial derivative.

Here is the chain rule for this partial derivative.

The two-variable chain rule leads to a formula that takes most of the work out of implicit differentiation. Suppose (1) the function F (x, y) is differentiable, and (2) the equation F (x, y) = 0 defines y implicitly as a differentiable function of x, say y = h (x).

EXAMPLE 7:

Assuming that the equation xy + y ² - 3x - 3 = 0 defines y as a differentiable function of x, use the above equation to find the value of dy/dx.

SOLUTION:

F = xy + y ² - 3x - 3 = 0

F_x = y - 3

F_y = x + 2y

EXAMPLE 8:

Assuming that the equation xe ^y + sin xy + y - ln 2 = 0 defines y as a differentiable function of x, use the above equation to find the value of dy/dx.

SOLUTION:

F = xe ^y + sin xy + y - ln 2 = 0

F_x = e ^y + y cos xy

F_y = xe ^y + x cos xy + 1