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PD.5 The Chain Rule

We recall 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


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However, there are many different ways of using the chain rule in equations that have more than two variables. This first example deals with the case where $z=f(x,y)$, and each of the variables x and y are in turn functions of a variable t. This means that the function looks like : $z=f(g(t),h(t)).$ This version of the chain rule is:

 

 

Case 1:

Suppose that $z=f(g(t),h(t))$, Then $x$ is a differentiable function of $t$ and:


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This can be rewritten as:


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Next we will consider the situation where $z=f(x,y)$ but $x$ and $y$ are a function of two variables $s$ and $t$: MATH This derivative can be found by doing the following:

 

Case 2:

Suppose that MATH Where $x=g(s,t)$ and $y=h(s,t)$ are differentiable functions of $s$ and $t.$Then:


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and


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In simple words, if you want to find MATH differentiate each equation in terms of $s$ while treating $t$ like a constant, or if you want to find MATH differentiate each equation in terms of $t$ while treating $s$ like a constant.

 

In Case 2 of the chain rule there are three different types of variables: $t$ and $s$ are the independent variables, $x$ and $y$ are called intermediate variables, and $z$ is the dependent variable.

 

To help us remember the chain rule it is helpful to draw a tree diagram. In this diagram we draw branches from the dependant variable z to the intermediate variables x and y to indicate the 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 derivatives. To find we find the product of the partial derivatives along each path from z to s and then add these products:



This tree diagram can be the following:


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Another example of a tree diagram of where $u=(x,y,z)$ and $x=x(r,s,t),$ $y=y(r,s,t),$ $z=z(r,s,t)$ is as follows:


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Now we will consider the general situation in which a dependant variable $u$ is a function of $n$ intermediate variables MATH, each of which is, in turn a function of $m$ variables $t_{1},....,t_{m}.$ Notice that there are $n$ terms, one for each intermediate variable. The formula is shown below:


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This says to differentiate each intermediate variable with respect to one independent variable while keeping the other variable as a constant.


 

Implicit Differentiation

 

For implicit differentiation, use the following equation:


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This formula is called the Implicit Function Theorem. Two other forms exist too, they are show below:


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and


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One can use these equations by simply solving the original equation for zero, and then find the partial derivatives, and then plug the partial derivatives into the formulas above.

 

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