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 $$4(1.001)^{5}-3(1.001)^{3}+2.$$

Solution

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