Derivatives of exponential and logarithmetic functions

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\todo }[0]{} \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\tkzTabSlope }[1]{\foreach \x /\y /\z in {#1}{\node [left = 3pt] at (Z\x 1) {\scriptsize $\y $}; \node [right = 3pt] at (Z\x 1) {\scriptsize $\z $}; }} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\ddt }[0]{\frac {d}{\d t}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{\vec {proj}} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[0]{\overrightarrow } \newcommand {\vec }[0]{\mathbf } \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\boldsymbol {\l }} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

Derivatives of exponential and logarithmic functions calculated.

The Derivative of the Natural Logarithm

We do not yet have a shortcut formula for the derivative of the natural logarithm, so let’s start from the definition. Set $f(x) = \ln x$ .

$\begin{align*} \dfrac{f(x+h) - f(x)}{h} &= \dfrac{\ln(x+h) - \ln(x)}{h} \\ &= \dfrac{ \ln \left( \dfrac{x+h}{x} \right) }{h} \\ &= \dfrac{1}{h} \cdot \ln \left( \dfrac{x+h}{x} \right) \\ &= \ln \left( \left( \dfrac{x+h}{x} \right)^{1/h} \right). \end{align*}$

That is, $f'(x) = \lim _{h \to 0} \ln \left ( \left ( \dfrac {x+h}{x} \right )^{1/h} \right ) = \ln \left ( \lim _{h \to 0} \left ( \dfrac {x+h}{x} \right )^{1/h} \right ).$ The limit inside the logarithm is a bit beyond what we can deal with right now, so unless we can come up with a different strategy, we’re stuck.

What do we know about this logarithm? We know that the natural logarithm function $f(x) = \ln x$ is the inverse of the exponential function $e^x$ That is, $y = \ln (x)\qquad \text {means}\qquad e^y = x.$ Since we’re trying to find the derivative of $\ln x$ , that means we’re trying to find $y'$ . Rather than working with the logarithmic version of $y = \ln x$ , let’s try to work with its exponential version $e^y = x$ . We’ll start by taking the derivative of both sides $\ddx \left ( e^y \right ) = \ddx \left ( x \right ).$ The right-hand side is easy, but what about the left-hand side? If we think of $y$ as just a function of $x$ , then the left-hand side is the exponential $e^x$ with the $x$ replaced by a function. It’s a Chain Rule problem, when we think of $e^y$ as having an outside function $g(x) = e^x$ and an inside function $f(x) = y$ . By Chain Rule, $\ddx g(f(x)) = g'(f(x)) f'(x) = e^y y'$ . Let’s put all this together.

$\begin{align*} e^y &= x\\ \ddx \left( e^y \right) &= \ddx \left( x \right) \\ e^y y' &= \answer[given]{1}\\ y' &= \dfrac{1}{e^y}. \end{align*}$

We notice once again that $e^y = x$ , so $\dfrac {1}{e^y} = \dfrac {1}{x}$ . This gives our derivative formula.

The derivative of the natural logarithm function The derivative of the natural logarithm function is given by $\ddx \ln (x) = \dfrac {1}{x}$ for $x > 0$ .

Compute $\displaystyle \ddx \ln \left (2x+1\right )$ .

This is a Chain Rule question with outer function $f(x) = \ln x$ and inner function $g(x) = 2x+1$ . We know $f'(x) = \dfrac {1}{x}$ and $g'(x) = \answer [given]{2}$ , so that $\ddx \ln \left ( 2x+1\right ) = \dfrac {\answer [given]{2}}{\answer [given]{2x+1}}.$

Compute $\displaystyle \ddx e^{3x} \ln \left ( 4x^2+3\right )$ .

$\displaystyle e^{3x} \dfrac {1}{4x^2+3}$ $\displaystyle 3e^{3x} \dfrac {8x}{4x^2+3}$ $\displaystyle 3e^{3x}\ln (4x^2+3) + e^{3x} \dfrac {8x}{4x^2+3}$ $\displaystyle 3e^{3x} + \dfrac {8x}{4x^2+3}$

The Derivative of Exponentials and Logarithms with Other Bases

We have found derivative formulas for the natural exponential function $e^x$ and the natural logarithm function $\ln x$ , but we have not yet explored other bases. That will be our focus for the rest of the section.

For exponentials, we remember that any number $b > 0$ can be written in the form $e^x$ for some specific value of $x$ . To determine the $x$ , we solve the equation $b = e^x$ so $x = \ln b$ . That is, $b = e^{\ln b}$ .

$\begin{align*} b^x &= (b)^x\\ &= \left( e^{\ln b} \right)^x\\ &= e^{x \ln b}. \end{align*}$

To find $\ddx b^x$ we are finding $\ddx e^{x \ln b}$ , which we know by Chain Rule.

$\begin{align*} \ddx b^x &= \ddx e^{x \ln b}\\ &= e^{x \ln b} \cdot \ddx \left( x \ln b \right)\\ &= e^{x \ln b} \cdot (\ln b). \end{align*}$

Rewriting $e^{x \ln b}$ as $b^x$ we find our derivative formula.

The derivative of the exponential function $b^x$ , for $b > 0$ , $b \neq 1$ is given by: $\ddx b^x = b^x \ln b.$

Compute $\displaystyle \ddx \sqrt {2}^x$ .

This is directly from our formula with $b = \sqrt {2}$ . We get $\displaystyle \sqrt {2}^x \ln \sqrt 2$ .

Compute $\displaystyle \ddx 3^{x^2+2x}$ .

This is the $b=3$ situation, but the $x$ has been replaced by $x^2 + 2x$ . That means we’ll need Chain Rule, too. $\begin{align*} \ddx 3^{x^2 + 2x} &= 3^{\answer[given]{x^2+2x}} \cdot \ln \left( \answer[given]{3} \right) \cdot \ddx \left( \answer[given]{x^2+2x}\right)\\ &= 3^{x^2+2x} \ln(3) (2x+2). \end{align*}$

To deal with logarithms of other bases, we rely on the change of base formula:

The change of base formula Let $b > 0$ with $b \neq 1$ . Then for any $a > 0$ and $a \neq 1$ and any $x > 0$ , $\log _b x = \dfrac {\log _a x }{ \log _a b }.$

This formula allows us to replace a logarithm with one base with a logarithm with whatever base we want. There is one base that we like more than the rest, base $e$ . This means $\displaystyle \log _b x = \dfrac { \ln x}{ \ln b}$ .

$\begin{align*} \ddx \log_b x &= \ddx \left( \dfrac{ \ln x }{\ln b} \right)\\ &= \dfrac{1}{\ln b} \cdot \ddx\left( \ln x \right) \\ &= \dfrac{1}{\ln b} \cdot \dfrac{1}{x}. \end{align*}$

This gives our derivative formula.

The derivative of the logarithmic function $\log _b x$ , for $b > 0$ , $b \neq 1$ is given by: $\ddx log_b x = \dfrac {1}{x \ln b}.$

Compute $\displaystyle \ddx \log _7(2x-3)$ .

This is the formula for $b=7$ , with $x$ replaced by $2x-3$ . We will need Chain Rule with inner function $g(x) = 2x-3$ and outer function $f(x) = \log _7 x$ . We know $g'(x) = \answer [given]{2}$ and $f'(x) = \dfrac {\answer [given]{1}}{x \ln \left ( \answer [given]{7} \right )}$ . Then The derivative is $\dfrac {2}{(2x-3) \ln 7}.$

Compute $\displaystyle \ddx 5^{2x} \log _{\frac {1}{3}}\left ( 6x^3-2x\right )$ .

The outer most operation in this function is multiplication, so this is a product rule question. We know the derivative of $5^{2x}$ is $5^{2x} \cdot 2$ , and the derivative of $\log _{\frac {1}{3}}\left (6x^3 - 2x \right )$ is $\displaystyle \dfrac {1}{(6x^3-2x) \ln \left (\answer [given]{\frac {1}{3}}\right )} \cdot \left (\answer [given]{18x^2-2}\right ))$ . The whole derivative is $2 \cdot 5^{2x}\cdot \log _{\frac {1}{3}}\left ( 6x^3-2x\right ) + 5^{2x}\cdot \dfrac {18x^2-2}{(6x^3-2x)\ln \left (\frac {1}{3}\right )} .$

Compute $\displaystyle \ddx 6^{x^2} \log _5\left (x+3\right )$ .

$\displaystyle 6^{2x} \dfrac {1}{(x+3) \ln 5}$ $\displaystyle 2x \cdot 6^{x^2}\ln (6)\log _5(x+3) + 6^{x^2} \dfrac {1}{(x+3) \ln 5}$ $\displaystyle x^2 \cdot 6^{x^2-1} \log _5(x+3) + 6^{x^2} \dfrac {1}{(x+3) \ln 5}$ $\displaystyle 2x \cdot 6^{x^2}\ln (6)\log _5(x+3) + 6^{x^2} \dfrac {5}{x+3}$