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4.5 The chain rule  (Page 2/9)

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Since x ( t ) and y ( t ) are both differentiable functions of t , both limits inside the last radical exist. Therefore, this value is finite. This proves the chain rule at t = t 0 ; the rest of the theorem follows from the assumption that all functions are differentiable over their entire domains.

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Closer examination of [link] reveals an interesting pattern. The first term in the equation is ∂ f ∂ x · d x d t and the second term is ∂ f ∂ y · d y d t . Recall that when multiplying fractions, cancelation can be used. If we treat these derivatives as fractions, then each product “simplifies” to something resembling ∂ f / d t . The variables x and y that disappear in this simplification are often called intermediate variables : they are independent variables for the function f , but are dependent variables for the variable t . Two terms appear on the right-hand side of the formula, and f is a function of two variables. This pattern works with functions of more than two variables as well, as we see later in this section.

Using the chain rule

Calculate d z / d t for each of the following functions:

  1. z = f ( x , y ) = 4 x 2 + 3 y 2 , x = x ( t ) = sin t , y = y ( t ) = cos t
  2. z = f ( x , y ) = x 2 − y 2 , x = x ( t ) = e 2 t , y = y ( t ) = e − t
  1. To use the chain rule, we need four quantities— ∂ z / ∂ x , ∂ z / ∂ y , d x / d t , and d y / d t :
    ∂ z ∂ x = 8 x ∂ z ∂ y = 6 y d x d t = cos t d y d t = − sin t

    Now, we substitute each of these into [link] :
    d z d t = ∂ z ∂ x · d x d t + ∂ z ∂ y · d y d t = ( 8 x ) ( cos t ) + ( 6 y ) ( − sin t ) = 8 x cos t − 6 y sin t .

    This answer has three variables in it. To reduce it to one variable, use the fact that x ( t ) = sin t and y ( t ) = cos t . We obtain
    d z d t = 8 x cos t − 6 y sin t = 8 ( sin t ) cos t − 6 ( cos t ) sin t = 2 sin t cos t .

    This derivative can also be calculated by first substituting x ( t ) and y ( t ) into f ( x , y ) , then differentiating with respect to t :
    z = f ( x , y ) = f ( x ( t ) , y ( t ) ) = 4 ( x ( t ) ) 2 + 3 ( y ( t ) ) 2 = 4 sin 2 t + 3 cos 2 t .

    Then
    d z d t = 2 ( 4 sin t ) ( cos t ) + 2 ( 3 cos t ) ( − sin t ) = 8 sin t cos t − 6 sin t cos t = 2 sin t cos t ,

    which is the same solution. However, it may not always be this easy to differentiate in this form.
  2. To use the chain rule, we again need four quantities— ∂ z / ∂ x , ∂ z / d y , d x / d t , and d y / d t :
    ∂ z ∂ x = x x 2 − y 2 ∂ z ∂ y = − y x 2 − y 2 d x d t = 2 e 2 t d x d t = − e − t .

    We substitute each of these into [link] :
    d z d t = ∂ z ∂ x · d x d t + ∂ z ∂ y · d y d t = ( x x 2 − y 2 ) ( 2 e 2 t ) + ( − y x 2 − y 2 ) ( − e − t ) = 2 x e 2 t − y e − t x 2 − y 2 .

    To reduce this to one variable, we use the fact that x ( t ) = e 2 t and y ( t ) = e − t . Therefore,
    d z d t = 2 x e 2 t + y e − t x 2 − y 2 = 2 ( e 2 t ) e 2 t + ( e − t ) e − t e 4 t − e −2 t = 2 e 4 t + e −2 t e 4 t − e −2 t .

    To eliminate negative exponents, we multiply the top by e 2 t and the bottom by e 4 t :
    d z d t = 2 e 4 t + e −2 t e 4 t − e −2 t · e 2 t e 4 t = 2 e 6 t + 1 e 8 t − e 2 t = 2 e 6 t + 1 e 2 t ( e 6 t − 1 ) = 2 e 6 t + 1 e t e 6 t − 1 .

    Again, this derivative can also be calculated by first substituting x ( t ) and y ( t ) into f ( x , y ) , then differentiating with respect to t :
    z = f ( x , y ) = f ( x ( t ) , y ( t ) ) = ( x ( t ) ) 2 − ( y ( t ) ) 2 = e 4 t − e −2 t = ( e 4 t − e −2 t ) 1 / 2 .

    Then
    d z d t = 1 2 ( e 4 t − e −2 t ) − 1 / 2 ( 4 e 4 t + 2 e −2 t ) = 2 e 4 t + e −2 t e 4 t − e −2 t .

    This is the same solution.
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Calculate d z / d t given the following functions. Express the final answer in terms of t .

z = f ( x , y ) = x 2 − 3 x y + 2 y 2 , x = x ( t ) = 3 sin 2 t , y = y ( t ) = 4 cos 2 t

d z d t = ∂ f ∂ x d x d t + ∂ f ∂ y d y d t = ( 2 x − 3 y ) ( 6 cos 2 t ) + ( −3 x + 4 y ) ( −8 sin 2 t ) = −92 sin 2 t cos 2 t − 72 ( cos 2 2 t − sin 2 2 t ) = −46 sin 4 t − 72 cos 4 t .

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It is often useful to create a visual representation of [link] for the chain rule. This is called a tree diagram    for the chain rule for functions of one variable and it provides a way to remember the formula ( [link] ). This diagram can be expanded for functions of more than one variable, as we shall see very shortly.
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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