Solving Differentials Problems

The following problems involve the concept of the Differential of a Function. I will introduce the Differential via it's geometric interpretation and then formulate it. For a function $ y=f(x) $ it will be shown that Differentials can be used to estimate the change of $y-$values as a function of a change in $x$-values. We will use Differentials to solve three types of problems. Differentials will be used to

$ \ \ \ \ \ $ I.) estimate the value of a numerical expression.
$ \ \ \ \ $ II.) approximate a relatively complicated functional expression with a simpler polynomial expression.
$ \ \ \ $ III.) estimate the propagation of percentage errors.

Let's begin with the graph of a function $y=f(x)$ and consider $x-$values changing from $x$ to $x+ \Delta x$, where $ \Delta x$ will be used to represent a small positive or negative change in $x$. I will refer to $x$ as the "starting" $x-$value and to $x + \Delta x$ as the second $x-$value. Draw a tangent line to this graph at $x$.

tex2html_wrap_inline125



Since $x-$values change from $x$ to $x+ \Delta x$, the corresponding $y-$values will change from $ f(x) $ to $ f(x + \Delta x) $. Define this exact change in $y$-values to be $ \Delta y $, where $$ \Delta y = f(x + \Delta x) - f(x) $$ We can now geometrically define this so-called Differential of $f(x)$. We will denote the Differential by $ dy $. It is the HEIGHT of the designated right triangle formed by $x$, $x+ \Delta x$, and the tangent line to the graph of $y=f(x)$ at $x$ in the following diagram:

tex2html_wrap_inline125



Let's find a formula for $ dy$. Recall from algebra that the SLOPE of this tangent line at $x$ is $$ m = \displaystyle{ rise \over run } = \displaystyle{ dy \over \Delta x } $$ Recall also that the SLOPE of this tangent line at $x$ is the derivative $$ m=f'(x) $$ Setting these slopes equal to each other gives us $$ \displaystyle{ dy \over \Delta x } = f'(x) $$

so that the Differential of Function $f$ at $x$ is

$$ dy = f'(x) \ \Delta x $$

Let's now verify that the Exact Change of $ \ y=f(x) \ $ is approximately equal to the Differential of $ \ y=f(x) \ $ for "small" $ \Delta x $, i.e.,

$$ \Delta y \approx dy $$

for small $ \Delta x $. Recall that the Derivative of $f$ at $x$ is $$ \displaystyle{ \lim_{ \Delta x \to 0 } { f(x + \Delta x) - f(x) \over \Delta x } } = f'(x) \ \ \ \ \longrightarrow $$ $$ \displaystyle{ { f(x + \Delta x) - f(x) \over \Delta x } } \approx f'(x) \ \ \ \ \longrightarrow $$
for "small" $ \Delta x$
$$ \displaystyle{ { \Delta y \over \Delta x } } \approx f'(x) \ \ \ \ \longrightarrow $$ $$ \Delta y \approx f'(x) \ \Delta x \ \ \ \ \longrightarrow $$ $$ \Delta y \approx dy $$
for "small" $ \Delta x $.

Here is a summary of Differentials facts.

$ \ \ \ \ $ 1. Differentials require a function, $ y=f(x)$.
$ \ \ \ \ $ 2. Differentials require two $x-$values, written as $ \ x-$values: $x \ \rightarrow x + \Delta x $, where $x$ is denoted as the "starting" $x-$value and $ \Delta x$ can be positive or negative.
$ \ \ \ \ $ 3. The Exact Change in $y-$values is $ \ \Delta y = f(x+ \Delta x)- f(x) $.
$ \ \ \ \ $ 4. The Differential formula is $ \ dy = f'(x) \ \Delta x $, where $x$ is the "starting" $x-$value.
$ \ \ \ \ $ 5. We will assume that $ \ \Delta y \approx dy \ $ if $ \Delta x $ is "small."


In the list of Differentials Problems which follows, most problems are average and a few are somewhat challenging.


CATEGORY I-- Using Differentials to Estimate the Value of a Numerical Expression



CATEGORY II-- Using Differentials to Approximate a Relatively Complicated Functional Expression with a Simpler Polynomial Expression



CATEGORY II-- Using Differentials to Approximate the Percentage Errors

For the following problems we will refer to $ |\Delta x| $ as the "absolute error in $x$ and to $ |\Delta y| $ as the "absolute error in $y$. We will define $ \displaystyle{ |\Delta x| \over x } $ to be the "absolute percentage error in $x$" and $ \displaystyle{ |\Delta y| \over y } $ to be the "absolute percentage error in $y$". For example, if $ \Delta x= -0.4 $ and $x=20$, then the absolute percentage error in $x$ is $$ \displaystyle{ |-0.4| \over 20 }= { 0.4 \over 20 } = { 0.4 \over 20 }{ 5 \over 5} = { 2 \over 100 } = 2\% $$

CATEGORY III-- Miscellaneous Differential Problems



Click HERE to return to the original list of various types of calculus problems.

Your comments and suggestions are welcome. Please e-mail any correspondence to Duane Kouba by clicking on the following address :

kouba@math.ucdavis.edu

A heartfelt "Thank you" goes to The MathJax Consortium for making the construction of this webpage fun and easy.

Duane Kouba ... October 24, 2019