4.5 The chain rule

Calculus volume 3 Page 1 / 9

State the chain rules for one or two independent variables.
Use tree diagrams as an aid to understanding the chain rule for several independent and intermediate variables.
Perform implicit differentiation of a function of two or more variables.

In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. The same thing is true for multivariable calculus, but this time we have to deal with more than one form of the chain rule. In this section, we study extensions of the chain rule and learn how to take derivatives of compositions of functions of more than one variable.

Chain rules for one or two independent variables

Recall that the chain rule for the derivative of a composite of two functions can be written in the form

\frac{d}{d x} (f (g (x))) = f' (g (x)) g' (x) .

In this equation, both $f (x)$ and $g (x)$ are functions of one variable. Now suppose that $f$ is a function of two variables and $g$ is a function of one variable. Or perhaps they are both functions of two variables, or even more. How would we calculate the derivative in these cases? The following theorem gives us the answer for the case of one independent variable.

Chain rule for one independent variable

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

\frac{d z}{d t} = \frac{\partial z}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial z}{\partial y} \cdot \frac{d y}{d t},

where the ordinary derivatives are evaluated at $t$ and the partial derivatives are evaluated at $(x, y) .$

Proof

The proof of this theorem uses the definition of differentiability of a function of two variables. Suppose that f is differentiable at the point $P (x_{0}, y_{0}),$ where $x_{0} = g (t_{0})$ and $y_{0} = h (t_{0})$ for a fixed value of $t_{0} .$ We wish to prove that $z = f (x (t), y (t))$ is differentiable at $t = t_{0}$ and that [link] holds at that point as well.

Since $f$ is differentiable at $P,$ we know that

z (t) = f (x, y) = f (x_{0}, y_{0}) + f_{x} (x_{0}, y_{0}) (x - x_{0}) + f_{y} (x_{0}, y_{0}) (y - y_{0}) + E (x, y),

where $lim_{(x, y) \to (x_{0}, y_{0})} \frac{E (x, y)}{\sqrt{{(x - x_{0})}^{2} + {(y - y_{0})}^{2}}} = 0 .$ We then subtract $z_{0} = f (x_{0}, y_{0})$ from both sides of this equation:

\begin{matrix} z (t) - z (t_{0}) & = f (x (t), y (t)) - f (x (t_{0}), y (t_{0})) \\ = f_{x} (x_{0}, y_{0}) (x (t) - x (t_{0})) + f_{y} (x_{0}, y_{0}) (y (t) - y (t_{0})) + E (x (t), y (t)) . \end{matrix}

Next, we divide both sides by $t - t_{0} :$

\frac{z (t) - z (t_{0})}{t - t_{0}} = f_{x} (x_{0}, y_{0}) (\frac{x (t) - x (t_{0})}{t - t_{0}}) + f_{y} (x_{0}, y_{0}) (\frac{y (t) - y (t_{0})}{t - t_{0}}) + \frac{E (x (t), y (t))}{t - t_{0}} .

Then we take the limit as $t$ approaches $t_{0} :$

\begin{matrix} lim_{t \to t_{0}} \frac{z (t) - z (t_{0})}{t - t_{0}} & = f_{x} (x_{0}, y_{0}) lim_{t \to t_{0}} (\frac{x (t) - x (t_{0})}{t - t_{0}}) + f_{y} (x_{0}, y_{0}) lim_{t \to t_{0}} (\frac{y (t) - y (t_{0})}{t - t_{0}}) \\ + lim_{t \to t_{0}} \frac{E (x (t), y (t))}{t - t_{0}} . \end{matrix}

The left-hand side of this equation is equal to $d z / d t,$ which leads to

\frac{d z}{d t} = f_{x} (x_{0}, y_{0}) \frac{d x}{d t} + f_{y} (x_{0}, y_{0}) \frac{d y}{d t} + lim_{t \to t_{0}} \frac{E (x (t), y (t))}{t - t_{0}} .

The last term can be rewritten as

\begin{matrix} lim_{t \to t_{0}} \frac{E (x (t), y (t))}{t - t_{0}} & = lim_{t \to t_{0}} (\frac{E (x, y)}{\sqrt{{(x - x_{0})}^{2} + {(y - y_{0})}^{2}}} \frac{\sqrt{{(x - x_{0})}^{2} + {(y - y_{0})}^{2}}}{t - t_{0}}) \\ = lim_{t \to t_{0}} (\frac{E (x, y)}{\sqrt{{(x - x_{0})}^{2} + {(y - y_{0})}^{2}}}) lim_{t \to t_{0}} (\frac{\sqrt{{(x - x_{0})}^{2} + {(y - y_{0})}^{2}}}{t - t_{0}}) . \end{matrix}

As $t$ approaches $t_{0},$ $(x (t), y (t))$ approaches $(x (t_{0}), y (t_{0})),$ so we can rewrite the last product as

lim_{(x, y) \to (x_{0}, y_{0})} (\frac{E (x, y)}{\sqrt{{(x - x_{0})}^{2} + {(y - y_{0})}^{2}}}) lim_{(x, y) \to (x_{0}, y_{0})} (\frac{\sqrt{{(x - x_{0})}^{2} + {(y - y_{0})}^{2}}}{t - t_{0}}) .

Since the first limit is equal to zero, we need only show that the second limit is finite:

\begin{matrix} lim_{(x, y) \to (x_{0}, y_{0})} (\frac{\sqrt{{(x - x_{0})}^{2} + {(y - y_{0})}^{2}}}{t - t_{0}}) & = lim_{(x, y) \to (x_{0}, y_{0})} (\sqrt{\frac{{(x - x_{0})}^{2} + {(y - y_{0})}^{2}}{{(t - t_{0})}^{2}}}) \\ = lim_{(x, y) \to (x_{0}, y_{0})} (\sqrt{{(\frac{x - x_{0}}{t - t_{0}})}^{2} + {(\frac{y - y_{0}}{t - t_{0}})}^{2}}) \\ = \sqrt{{(lim_{(x, y) \to (x_{0}, y_{0})} (\frac{x - x_{0}}{t - t_{0}}))}^{2} + {(lim_{(x, y) \to (x_{0}, y_{0})} (\frac{y - y_{0}}{t - t_{0}}))}^{2}} . \end{matrix}

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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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