**14.5** The Chain Rule

### The Chain Rule

We know that the Chain Rule for functions of a single
variable gives the rule for differentiating a composite
*function: If y = f(x) and x = g(t), where f and g are *

*differentiable functions, then y is indirectly a differentiable *
*function of t and*

For functions of more than one variable, the Chain Rule has several versions, each of them giving a rule for

differentiating a composite function.

### The Chain Rule

The first version (Theorem 2) deals with the case where
*z = f(x, y) and each of the variables x and y is, in turn, a *

*function of a variable t.*

*This means that z is indirectly a function of t, *

*z = f(g(t), h(t)), and the Chain Rule gives a formula for *
*differentiating z as a function of t. We assume that f is *
differentiable.

### The Chain Rule

*We know that this is the case when f*_{x}*and f** _{y}* are continuous.

Since we often write *∂z/∂x in place of ∂f/∂x, we can rewrite*
the Chain Rule in the form

### Example 1

*If z = x*^{2}*y + 3xy*^{4}*, where x = sin 2t and y = cos t, find dz/dt*
*when t = 0.*

Solution:

The Chain Rule gives

*It’s not necessary to substitute the expressions for x and y*
*in terms of t.*

*Example 1 – Solution*

*We simply observe that when t = 0, we have x = sin 0 = 0 *
*and y = cos 0 = 1. *

Therefore

cont’d

### The Chain Rule

*We now consider the situation where z = f(x, y) but each of *
*x and y is a function of two variables s and t: *

*x = g(s, t), y = h(s, t). *

*Then z is indirectly a function of s and t and we wish to find *

*∂z/∂s and ∂z/∂t. *

We know that in computing *∂z/∂t we hold s fixed and *
*compute the ordinary derivative of z with respect to t. *

Therefore we can apply Theorem 2 to obtain

### The Chain Rule

A similar argument holds for *∂z/∂s and so we have proved *
the following version of the Chain Rule.

Case 2 of the Chain Rule contains three types of variables:

**s and t are independent variables, x and y are called ****intermediate variables, and z is the dependent variable.**

### The Chain Rule

Notice that Theorem 3 has one term for each intermediate variable and each of these terms resembles the one-dimensional Chain Rule in Equation 1.

**To remember the Chain Rule, it’s helpful to draw the tree **
**diagram in Figure 2.**

### The Chain Rule

*We draw branches from the dependent variable z to the *
*intermediate variables x and y to indicate that z is a *

*function of x and y. Then we draw branches from x and y to *
*the independent variables s and t. *

On each branch we write the corresponding partial

derivative. To find *∂z/∂s, we find the product of the partial *
*derivatives along each path from z to s and then add these *
products:

### The Chain Rule

Similarly, we find *∂z/∂t by using the paths from z to t.*

Now we consider the general situation in which a
*dependent variable u is a function of n intermediate *

*variables x*_{1}*, …, x*_{n}*, each of which is, in turn, a function of m*
*independent variables t*_{1}*,…, t** _{m}*.

*Notice that there are n terms, one for each intermediate *
variable. The proof is similar to that of Case 1.

### The Chain Rule

### Implicit Differentiation

### Implicit Differentiation

The Chain Rule can be used to give a more complete description of the process of implicit differentiation.

*We suppose that an equation of the form F(x, y) = 0 *

*defines y implicitly as a differentiable function of x, that is, *
*y = f(x), where F(x, f(x)) = 0 for all x in the domain of f. *

*If F is differentiable, we can apply Case 1 of the Chain Rule *
*to differentiate both sides of the equation F(x, y) = 0 with *
*respect to x. *

*Since both x and y are functions of x, we obtain *

### Implicit Differentiation

*But dx/dx = 1, so if ∂F/∂x ≠ 0 we solve for dy/dx and obtain*

*To derive this equation we assumed that F(x, y) = 0 defines*
*y implicitly as a function of x. *

### Implicit Differentiation

**The Implicit Function Theorem, proved in advanced **

calculus, gives conditions under which this assumption is
*valid: it states that if F is defined on a disk containing (a, b), *
*where F(a, b) = 0, F*_{y}*(a, b) ≠ 0, and F*_{x}*and F** _{y}* are

*continuous on the disk, then the equation F(x, y) = 0 *
*defines y as a function of x near the point (a, b) and the *
derivative of this function is given by Equation 6.

### Example 8

*Find y′ if x*^{3} *+ y*^{3} *= 6xy.*

Solution:

The given equation can be written as
*F(x, y) = x*^{3} *+ y*^{3} *– 6xy = 0*
so Equation 6 gives

### Implicit Differentiation

*Now we suppose that z is given implicitly as a function *
*z = f(x, y) by an equation of the form F(x, y, z) = 0. *

*This means that F(x, y, f(x, y)) = 0 for all (x, y) in the *

*domain of f. If F and f are differentiable, then we can use *
*the Chain Rule to differentiate the equation F(x, y, z) = 0 as *
follows:

### Implicit Differentiation

But and

so this equation becomes

If *∂F/∂z ≠ 0, we solve for ∂z/∂x and obtain the first formula *
in Equations 7.

The formula for *∂z/∂y is obtained in a similar manner.*

### Implicit Differentiation

**Again, a version of the Implicit Function Theorem **

stipulates conditions under which our assumption is valid:

*if F is defined within a sphere containing (a, b, c), where *
*F(a, b, c) = 0, F*_{z}*(a, b, c) ≠ 0, and F*_{x}*, F*_{y}*, and F** _{z}* are

continuous inside the sphere, then the equation

*F(x, y, z) = 0 defines z as a function of x and y near the*
*point (a, b, c) and this function is differentiable, with partial*