Short and Sweet Calculus

3.13 Differentials

At the beginning of this chapter, we used the symbol \(dy/dx\) as a single entity to denote the limit of the quotient \(\Delta y/\Delta x\) as \(\Delta x\to0\). In many cases, it is desirable to consider \(dx\) and \(dy\) as separate quantities. To this end, we need to define \(dx\) and \(dy\) in such a manner that the derivative \(dy/dx=f'(x)\) will be actually the ordinary quotient of \(dy\) and \(dx\).

Let \(dx\) be identical with the increment of \(x\); that is, \(dx=\Delta x\). Since \(dx\) may assume any arbitrary value, \(dx\) is independent of \(x\) and any other quantity.

Now \(dx\) being defined, we need to define \(dy\) such that its quotient by \(dx\) is \(f'(x)\). So, we have \[\boxed{dy=f'(x)dx.}\] Here \(dy\) is a dependent variable whose value depends both on \(x\) and \(dx\).

If \(\Delta x\) is small and \(dx=\Delta x\), then \(dy\) is a good approximation for \(\Delta y\) (see Figure 3.10): \[\Delta y\approx dy\] or \[\underbrace{f(x_{0}+dx)-f(x_{0})}_{\Delta y}\approx\underbrace{f'(x_{0})dx.}_{dy}\] This is exactly the linear approximation of \(f\) (that we learned in the previous section) using the notation of differentials.

Example 3.25. Use differentials to estimate \(4(1.001)^{5}-3(1.001)^{3}+2.\)


When \(x=1\), the evaluation of \(f(x)=4x^{5}-3x^{3}+2\) is easy \[f(1)=4-3+2=3,\] so we take \(f(x)=4x^{4}-3x^{3}+2\) and \(dx=0.001\). Since \(dy=f'(x)dx=\left(20x^{4}-9x^{2}\right)dx\), for \(x=1\) and \(dx=0.001\) we have \[dy=(20-9)(0.001)=11\times0.001=0.011.\] Therefore, \[f(1.001)\approx f(1)+dy=3+0.011=3.011.\] The acutal value of \(f(1.001)\) to five decimals is 3.01103.

Properties of differentials

We may multiply both sides of the familiar formulas for evaluating a derivative and obtain corresponding formulas for differentials.

Derivative Formula Differential Formula
\(\dfrac{d}{dx}c=0\quad\) (\(c\) is a constant) \(dc=0\)
\(\dfrac{d}{dx}(cu)=c\dfrac{du}{dx}\) \(d(cu)=cdu\)
\(\dfrac{d}{dx}(u+v)=\dfrac{du}{dx}+\dfrac{dv}{dx}\) \(d(u+v)=du+dv\)
\(\dfrac{d}{dx}(uv)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\) \(d(uv)=udv+vdu\)
\(\dfrac{d}{dx}\left(\dfrac{u}{v}\right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{v^{2}}\) \(d\left(\dfrac{u}{v}\right)=\dfrac{vdu-udv}{v^{2}}\)
\(\dfrac{d}{dx}(x^{r})=rx^{r-1}\quad(r\in\mathbb{R})\) \(d(x^{r})=rx^{r-1}dx\)

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