MA005: Calculus I

The Chain Rule is the most important and most used of the differentiation patterns.  It enables us to differentiate composites of functions such as  $y = sin( x^2 )$.  It is a powerful tool for determining the derivatives of some new functions such as logarithms and inverse trigonometric functions.  And it leads to important applications in a variety of fields.  You  will need the Chain Rule hundreds of times in this course, and practice with it now will save you time and points later.   Fortunately, with some practice, the Chain Rule is also easy to use. 

We already know how to differentiate the composition of some functions.

Example 1: $\quad$ For $\mathrm{f}(\mathrm{x})=5 \mathrm{x}-4$ and $\mathrm{g}(\mathrm{x})=2 \mathrm{x}+1$, find $\operatorname{f \circ g}(\mathrm{x})$ and $\mathrm{D}(\operatorname{f \circ g}(\mathrm{x}))$

Solution: $\operatorname{f \circ g}(x)=f(g(x))=5(2 x+1)-4=10 x+1$, so $D(\operatorname{f \circ g}(x))=D(10 x+1)=10$.

Practice 1: For $\mathrm{f}(\mathrm{x})=5 \mathrm{x}-4$ and $\mathrm{g}(\mathrm{x})=\mathrm{x}^{2}$, find $\operatorname{f \circ g}(\mathrm{x}), \mathrm{D}(\operatorname{f \circ g}(\mathrm{x})), \mathrm{g} \circ \mathrm{f}(\mathrm{x})$, and $\mathrm{D}(\mathrm{g} \circ \mathrm{f}(\mathrm{x}))$.

Some compositions, however, are still very difficult to differentiate. We know the derivatives of $g(x)=x^{2}$ and $\mathrm{h}(\mathrm{x})=\sin (\mathrm{x})$, and we know how to differentiate some combinations of these functions such as $\mathrm{x}^{2}+\sin (\mathrm{x})$, $x^{2} \cdot \sin (x)$, and even $\sin ^{2}(x)$, but the derivative of the simple composition $f(x)=\operatorname{h \circ g}(x)=\sin \left(x^{2}\right)$ is hard - until we know the Chain Rule. To see just how hard, try using the definition of derivative on it.

Example 2:

(a) Suppose amplifier $\mathrm{Y}$ doubles the strength of the output signal from amplifier $\mathrm{U}$, and $\mathrm{U}$ triples the strength of the original signal $\mathrm{x} .$ How does the final signal out of $\mathrm{Y}$ compare with the signal $\mathrm{x}$ ?

original signal $\mathrm{x} \rightarrow$ amplifier $\mathrm{U} \rightarrow$ amplifier $Y$ $\rightarrow$ final signal

(b) Suppose $y$ changes twice as fast as $\mathrm{u}$, and $\mathrm{u}$ changes three times as fast as $\mathrm{x}.$ How does the rate of change of $y$ compare with the rate of change of $x$ ?

Solution: In each case we are comparing the result of a composition, and the answer to each question is 6 , the product of the two amplifications or rates of change.

In (a), we have that $\frac{\text { signal out of } \mathrm{Y}}{\text { signal } \mathrm{x}}=\frac{\text { signal out of Y }}{\text { signal out of } \mathrm{U}} \cdot \frac{\text { signal out of } \mathrm{U}}{\text { signal } \mathrm{x}}=(2)(3)=6$

$\operatorname{In}(b), \frac{\Delta y}{\Delta x}=\frac{\Delta y}{\Delta u} \cdot \frac{\Delta u}{\Delta x}=(2)(3)=6$

These examples are simple cases of the Chain Rule for differentiating a composition of functions.

Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-3.5-Chain-Rule.pdf
This work is licensed under a Creative Commons Attribution 3.0 License.