# 14.5 The Chain Rule

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## 14.5 The Chain Rule

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### 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.

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### 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.

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### The Chain Rule

We know that this is the case when fx and fy are continuous.

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

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### Example 1

If z = x2y + 3xy4, 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.

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### Example 1 – Solution

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

Therefore

cont’d

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### 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

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### 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.

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### 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.

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### 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:

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### 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 x1, …, xn, each of which is, in turn, a function of m independent variables t1,…, tm.

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

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### 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

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### 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.

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### 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, Fy(a, b) ≠ 0, and Fx and Fy 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.

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### Example 8

Find y′ if x3 + y3 = 6xy.

Solution:

The given equation can be written as F(x, y) = x3 + y3 – 6xy = 0 so Equation 6 gives

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### 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:

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### 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.

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### 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, Fz(a, b, c) ≠ 0, and Fx, Fy, and Fz 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

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