Differentials

A few days ago was I wrote about Linear Approximations. These are used, as the name implies, to approximate the value of a function, usually a complicated function, by way of using a much simpler linear function, as long as \(\boldsymbol{\Delta x}\) (the change in x) is small. Along with Linear Approximation, another topic covered in textbooks at about the same time is Differentials.

We can start with a simple graph of a function \(\boldsymbol{y = f\left(x\right)}\). It shows \(\boldsymbol{\Delta x \text{ (the change in } x\text{) and } \Delta y\text{ (the change in } y\text{).}}\) A linear approximation \(\boldsymbol{\color{red}{\left(\,y = L\left(x\right)\,\right)}}\) has been added to the second graph at \(\boldsymbol{ \left(\,c , f\left(c\right)\,\right)}\). This second graph also shows \(\boldsymbol{\color{red}{dy}\; \text{(the differential of } \color{red}{y} \text{), as well as }\color{red}{dx}}\). Notice that \(\boldsymbol{\color{red}{dx} \text{ is the same thing as }\color{red}{\Delta x}}\).

Before continuing it is important to realize that in these examples the points chosen are far apart so that the labels can be shown clearly. In reality, differentials, as well as linear approximations, are useful only when \(\boldsymbol{\Delta x}\) is very small.

How do we obtain \(\boldsymbol{\color{red}{dy}}\)? Remember that linear approximation says that a function \(\boldsymbol{f}\) can be approximated as:

\begin{align*}
f\left(x\right) &\approx f\left(c\right) + f’\left(c\right) \left(x\;-\;c\right) \\
\text{First, re-write it: } f\left(x\right)\,- f\left(c\right) &\approx f’\left(c\right) \left(x\;-\;c\right) \\
\text{As the graphs show: } f\left(x\right)\,- f\left(c\right) &= \Delta y \quad \text{ and } \quad \left(x\;-\;c\right) = \Delta x \\
\text{We then have: } \quad \Delta y &\approx f’\left(c\right) \Delta x \\
\text{Which is the differential: } \quad dy &= f’\left(c\right) dx
\end{align*}

A simple memory aid is to recall if you have a function \(\boldsymbol{y = f\left(x\right)}\), its derivative notation is \(\boldsymbol{\frac{dy}{dx} = f’\left(x\right)}\). If you think of \(\boldsymbol{\frac{dy}{dx}}\) as a regular fraction, using algebra you can re-write this equation as \(\boldsymbol{dy = f’\left(x\right)\;dx}\).

Example One

Let’s use the notation of differentials to write the approximate change in the function \(\boldsymbol{f\left(x\right) = 3 \cos^{2}x}\) given a small change in \(\boldsymbol{dx}\).

\begin{align*}
\text{With our function: }\quad f\left(x\right) &= 3 \cos^{2}x \\
\text{we have the derivative:}\quad f’\left(x\right) &= -6 \cos x \sin x \\
\text{using trig identities, it is simplified to: }\quad \qquad &= -3 \sin 2x \\
\text{then the equation for the differential is: } \qquad dy &= \left(-3 \sin 2x \right) dx
\end{align*}

Now, let’s use it. We know that…

\begin{align*}
\text{our function:} \quad f\left(\frac{\pi}{4}\right) &= 1.500000000 \\
\text{and its derivative: }\, f’\left(\frac{\pi}{4}\right) &= -3 \\
\text{We want to know the function value at } \left(\frac{\pi}{4} + 0.1\right) \text{That makes } dx &= 0.1 \\
\text{Using differentials, we have: }\quad dy &= \left(-3\right) \left(0.1\right) \\
\qquad dy &= -0.3
\end{align*}

What have we learned?

\begin{align*}
\text{Using differentials, the value at } \left(\frac{\pi}{4} + 0.1\right) \text{ is } f\left(\frac{\pi}{4}\right) + dy &= 1.500000000 + \left(-0.3\right) \\
&= 1.200000000 \\
\text{The exact value, using a graphing utility: } f\left(\frac{\pi}{4} + 0.1\right) &= 1.201996004 \\
\text{If we want the function value at } \left(\frac{\pi}{4} + 0.11\right) \text{, then } dy &= \left(-3\right) \left(0.11\right) \\
&= -0.33 \\
\text{In this case } f\left(\frac{\pi}{4}\right) + dy &= 1.500000000 + \left(-0.33\right) \\
&= 1.170000000 \\
\text{The exact value, using a graphing utility: } f\left(\frac{\pi}{4} + 0.11\right) &= 1.172655565
\end{align*}

Example Two

We know that \(\boldsymbol{\sqrt{16} = 4}\). Suppose we want to get the value of \(\boldsymbol{\sqrt{16.5}}\)? First, we have to be clear on some things:

\begin{align*}
\text{The function is: } \, f\left(x\right) &= \sqrt{x} \\
\text{Its derivative at } f’\left(16\right) &= \frac{1}{2 \sqrt{16}} = \frac{1}{8} \\
\text{and}\qquad dx &= 0.5 \\
\text{Putting everything together: } dy &= f’\left(x\right) dx \\
&= \left(\frac{1}{8}\right) \left(0.5\right) \\
&= 0.062500000 \\
\text{Consequently, } \quad f\left(16.5\right) &\approx 4.062500000 \\
\text{Using a graphing utility: } \quad f\left(16.5\right) &= 4.062019202
\end{align*}