Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 4

Differentiation

The Derivative Function

The concept of the derivative evolved through a long process, and in part independently, in astronomy, physics and other fields, and has important applications in each of these fields. Not surprisingly, different notations for the derivative developed in each field. Some notations have been superseded by more popular notations; others have remained. In this course, we use the most common notations interchangeably.

For the derivative at a number $a$, we have

${\displaystyle f^{\prime }\left(a\right)=\frac{d}{dx}f\left(x\right){|}_{x=a}=\frac{df}{dx}{|}_{x=a}=D_{x}f\left(a\right)\text{.}}$

For the derivative function, we have

${\displaystyle f^{\prime }\left(x\right)=\frac{d}{dx}f\left(x\right)=\frac{df}{dx}=D_{x}f\left(x\right)\text{.}}$

If $y=f\left(x\right)$, then we write

${\displaystyle f^{\prime }\left(a\right)=y^{\prime }\left(a\right)=\frac{dy}{dx}{|}_{x=a}\text{ and }{f}^{\prime }\left(x\right)=y^{\prime }=y^{\prime }\left(x\right)=\frac{dy}{dx}\text{.}}$

The notation ${\displaystyle \frac{dy}{dx}=\frac{df}{dx} }$ is the Leibnitz notation for the derivative.

Example 4.18.

1. By Example 4.14, the derivative function of $f\left(x\right)=\sqrt{x}$ is

   ${\displaystyle f^{\prime }\left(x\right)=\frac{1}{2\sqrt{x}}\text{.}}$

   We can also write

   ${\displaystyle \frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}\text{ and }\frac{d}{dx}\sqrt{x}{|}_{x=3}=\frac{1}{2\sqrt{x}}{|}_{x=3}=\frac{1}{2\sqrt{3}}\text{.}}$

2. By Example 4.15, the derivative function of ${\displaystyle f\left(x\right)=\frac{1}{x+3}}$ is

   ${\displaystyle f^{\prime }\left(x\right)=D_{x}f\left(x\right)=-\frac{1}{{\left(x+3\right)}^2}}$

   or

   ${\displaystyle D_x\frac{1}{x+3}=-\frac{1}{{\left(x+3\right)}^2}\text{;}}$

   therefore,

   ${\displaystyle D_{x}f\left(2\right)=-\frac{1}{{\left(2+3\right)}^2}=-\frac{1}{25}\text{.}}$

3. By Example 4.16, the derivative function of $f\left(x\right)=x^3$ is

   $f^{\prime }\left(x\right)=3x^2\text{,}\text{ and }{f}^{\prime }\left(3\right)=3\left(9\right)=27\text{.}$

Example 4.19. To apply Definition 4.5, we must understand how $f\left(x+h\right)$ is obtained.

Note that $x+h$ takes the place of the independent variable $x$. Hence, if $f\left(x\right)=x^2+x$, then

$f\left(x+h\right)={\left(x+h\right)}^2+\left(x+h\right)=x^2+2xh+h^2+x+h\text{.}$

The derivative function is

${\displaystyle \begin{array}{ll}\hfill f^{\prime }\left(x\right)& =\underset{h\to 0}{\lim }{\displaystyle \frac{f\left(x+h\right)-f\left(x\right)}{h}}\hfill \\ \hfill & =\underset{h\to 0}{\lim }{\displaystyle \frac{\left(x^2+2xh+h^2+x+h\right)-\left(x^2+x\right)}{h}}\hfill \\ \hfill & =\underset{h\to 0}{\lim }{\displaystyle \frac{2xh+h^2+h}{h}}\hfill \\ \hfill & =\underset{h\to 0}{\lim }\left(2x+h+1\right)=2x+1\text{.}\hfill \end{array}}$

That is, $f^{\prime }\left(x\right)=2x+1$.

As we note above, the line closest to a curve at a point $p$ is the tangent line that lies on $p$. This line is very useful for estimating values of functions.

Example 4.20. In Figure 4.16, below, we have the graph of the function $f\left(x\right)=\sqrt{x}$, and we can see that the tangent line near the point $\left(1,f\left(1\right)\right)=\left(1,1\right)$ is close to the curve.

Figure 4.16. Graph of the function $f\left(x\right)=\sqrt{x}$

Since $f\left(1\right)=1$, the value of the function near $1$ is approximately equal to $1$, that is $f\left(1\text{.}25\right)\approx f\left(1\right)$ and $f\left(0\text{.}75\right)\approx f\left(1\right)$; thus, $\sqrt{1\text{.}25}\approx 1$ and $\sqrt{0\text{.}75}\approx 1$.

To give a first approximation of these values near $1$, we use the equation of the tangent line at this point.

Let us assume that the tangent line has the equation

${\displaystyle y=\frac{x}{2}+\frac{1}{2}}$

(we will see why later on).

Then, from Figure 4.16, we conclude that $f\left(1\text{.}25\right)\approx y\left(1\text{.}25\right)$. So,

${\displaystyle f\left(1\text{.}25\right)\approx \frac{1\text{.}25}{2}+\frac{1}{2}=\frac{9}{8}\text{;}}$

that is, $f\left(1\text{.}25\right)=\sqrt{1\text{.}25}\approx 1\text{.}125$.

Similarly,

${\displaystyle f\left(0\text{.}75\right)\approx y\left(0\text{.}75\right)=\frac{0\text{.}75}{2}+\frac{1}{2}=0\text{.}875\text{;}}$

that is, $f\left(0\text{.}75\right)=\sqrt{0\text{.}75}\approx 0\text{.}875$. Use your calculator to check these square roots. You will see that, at least to the first decimal place, each of these approximations is correct.

Now, the slope of the tangent line at $\left(1,1\right)$ is given by the derivative at $1$. From Example 4.14, we have

${\displaystyle f^{\prime }\left(1\right)=\frac{1}{2\sqrt{1}}=\frac{1}{2}\text{.}}$

The equation of a line with slope $1∕2$ running through the point $\left(1,1\right)$ is

${\displaystyle y-1=\frac{x-1}{2}\text{ or }y=\frac{x}{2}+\frac{1}{2}\text{.}}$

Example 4.21. By Example 4.15, we have

${\displaystyle \frac{d}{dx}\frac{1}{x+3}=-\frac{1}{{\left(x+3\right)}^2}\text{.}}$

The slope of the tangent line at the point

${\displaystyle \left(1,f\left(1\right)\right)=\left(1\text{,}\frac{1}{1+3}\right)=\left(1\text{,}\frac{1}{4}\right)}$

is

${\displaystyle -\frac{1}{{\left(1+3\right)}^2}=-\frac{1}{16}}$

and the equation of the tangent line is

${\displaystyle y-\frac{1}{4}=-\frac{x-1}{16}\text{ or }y=-\frac{x}{16}+\frac{5}{16}\text{.}}$

Thus,

${\displaystyle f\left(x\right)\approx -\frac{x}{16}+\frac{5}{16}}$

for any $x$ near $1$. For example, for $x=1\text{.}1$, we find that

${\displaystyle \frac{1}{4\text{.}1}=f\left(1\text{.}1\right)\approx -\frac{1\text{.}1}{16}+\frac{5}{16}=\frac{39}{160}\text{ [Why?]}}$

Use your calculator to determine that $39∕160$ is in fact a good approximation of $1∕4\text{.}1\text{.}$

In general, we have the following proposition about the equation of the tangent line.

Proposition 4.6. If the function $f$ is continuous around $a$, then the equation of the tangent line at the point $\left(a,f\left(a\right)\right)$ is

$y-f\left(a\right)=f^{\prime }\left(a\right)\left(x-a\right)\text{ or }y=f^{\prime }\left(a\right)\left(x-a\right)+f\left(a\right)\text{,}$

and for any $b$ close to $a\text{,}$

$f\left(b\right)\approx f^{\prime }\left(a\right)\left(b-a\right)+f\left(a\right)\text{.}$