2.3 Derivatives of Natural Exponential and Log

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In this section we compute derivatives involving $e^x$ and $\ln (x)$ .

We begin by computing the derivative of the exponential function $f(x) = e^x$ . Recall that $e$ is a number whose value is approximately 2.72. Like $\pi$ and $\sqrt 2$ , $e$ is an irrational number, meaning that its decimal representation is non-terminating and non-repeating. The significance of the number $e$ in mathematics is underscored by the simplicity of the differentiation formula we are about to discover. The result hinges on a limit that we analyzed in the Numerical Limits section: $\lim _{x \to 0} \frac {e^x - 1}{x} = 1.$

The derivative of $f(x) = e^x$ is found as follows:

$\begin {aligned} f'(x) &= \lim _{h \to 0} \frac {f(x+h)-f(x)}{h} \\[5pt] &= \lim _{h \to 0}\frac {e^{x+h}-e^x}{h}\\[5pt] &= \lim _{h \to 0} \frac {e^x e^h-e^x}{h}\\[5pt] &= \lim _{h \to 0} \frac {e^x (e^h-1)}{h}\\[5pt] &= e^x \cdot 1 = e^x.\\[-5pt] \end {aligned}$

Thus the derivative of $e^x$ is itself!

Find the equation of the tangent line to the graph of $f(x) = e^x$ at $x = 0$ .

Use the derivative to find the slope, $m$

The point of tangency is $(0, f(0))$

Use the point slope form: $y-y_0 = m(x-x_0)$

The equation of the tangent line is $y = \answer {x+1}$

Find the equation of the tangent line to the graph of $f(x) = e^x$ at $x = 1$ .

Use the derivative to find the slope, $m$

The point of tangency is $(1, f(1))$

Use the point slope form: $y-y_0 = m(x-x_0)$

The equation of the tangent line is $y = \answer {ex}$

Find the following derivatives involving $e^x$ :

$\frac {d}{dx} \left (5e^x\right ) = \answer {5e^x}.$
$\frac {d}{dx} \left (-2e^x\right ) = \answer {-2e^x}.$
$\frac {d}{dx} \left (\frac {e^x}{2}\right ) = \answer {\frac {e^x}{2}}.$
$\frac {d}{dx} \left (x^2 + e^x\right ) = \answer {2x + e^x}.$
$\frac {d}{dx} \left (\sqrt x - 3e^x\right ) = \answer {(1/2)x^{-1/2} - 3e^x}.$
$\frac {d}{dx} \left (4e^x + \frac {1}{x^2}\right ) = \answer {4e^x -\frac {2}{x^3}}.$

Next, we use the definition of the derivative to find the derivative of the natural logarithm.

We will need the special limit

$\lim _{x \to 0} \frac {\ln (1 + x)}{x} = 1.$

We can use a table of values to verify this limit: $\begin {array}{ c | c | c | c | c } x & 0.1 & 0.01 & 0.001 & 0.0001 \\ \hline \frac {\ln (1+x)}{x} & 0.953 & 0.995 & 0.9995 & 0.99995 \end {array}$ and $\begin {array}{ c | c | c | c | c } x & -0.1 & -0.01 & -0.001 & -0.0001 \\ \hline \frac {\ln (1+x)}{x} & 1.054 & 1.005 & 1.0005 & 1.00005 \end {array}$ Now we are prepared to use the definition to find the derivative of $\ln (x)$ :

If $f(x) = \ln (x)$ then

$\begin {aligned} f'(x) &= \lim _{h \to 0} \frac {f(x+h)-f(x)}{h}\\[5pt] &= \lim _{h \to 0}\frac {\ln (x+h)-\ln (x)}{h}\\[5pt] &= \lim _{h \to 0} \frac {\ln (\frac {x+h}{x})}{h}\\[5pt] &= \lim _{h \to 0}\frac {\ln (1 + \frac {h}{x})}{h}\\[5pt] &= \lim _{k \to 0} \frac {\ln (1 + k)}{kx} \\[5pt] &= \frac {1}{x}. \end {aligned}$

Note that we made the substitution $k =\frac {h}{x}$ so that $h = kx$ and also note that $h\to 0$ is equivalent to $k\to 0$ . To recap, $\frac {d}{dx} \ln (x) = \frac {1}{x}$ which gives a memorable mathematical relationship between the transcendental natural logarithm function the rational reciprocal function.

If $f(x) = \frac {\ln (x)}{\ln (4)}$ then we use the constant multiple rule with $c = \frac {1}{\ln (4)}$ and we get $f'(x) = \tfrac {1}{\ln (4)} \cdot \frac {1}{x} = \frac {1}{x\ln (4)}.$

Find the following derivatives involving $\ln (x)$ :

$\frac {d}{dx} \left (5\ln (x)\right ) = \answer {5/x}.$
$\frac {d}{dx} \left (-2\ln (x)\right ) = \answer {-2/x}.$
$\frac {d}{dx} \left (\frac {\ln (x)}{2}\right ) = \answer {1/(2x)}.$
$\frac {d}{dx} \left (x^3 + \ln (x)\right ) = \answer {3x^2 + 1/x}.$
$\frac {d}{dx} \left (\sqrt x - 3\ln (x)\right ) = \answer {(1/2)x^{-1/2} - 3/x}.$
$\frac {d}{dx} \left (4\ln (x) + \frac {1}{x^2}\right ) = \answer {4/x -\frac {2}{x^3}}.$

In the formula $\tfrac {d}{dx} \ln (x) = \tfrac {1}{x}$ , the domains of the functions do not match. The domain of $1/x$ is all non-zero $x$ , but the domain of $\ln (x)$ is only $x >0$ . This leads us to ask if there is a function defined for $x<0$ whose derivative is $1/x$ . The answer is yes and the function is $\ln |x|$ . In other words, $\dd {x} \ln |x| = \frac {1}{x}.$ This result follows from the definition of the derivative exactly as it was used above. The absolute value bars will eventually drop in the computation because $1+h/x$ becomes positive as $h \to 0$ .

Below is a graph of $f(x) = \ln |x|$ (in blue) and its derivative, $f'(x) = 1/x$ (in purple). Notice that the slope of the tangent line to $f(x)$ (in red) is the height of the corresponding point on $f'(x)$ .