2.8 Derivatives of General Exponential and Log

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\ffrac }[2]{\frac {\text {\footnotesize $#1$}}{\text {\footnotesize $#2$}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\C }[0]{\mathbb C} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\dis }[0]{\displaystyle } \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\ppx }[0]{\frac {\partial }{\partial x}} \newcommand {\ppy }[0]{\frac {\partial }{\partial y}} \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]{ #1 \bigg |} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\proj }{proj} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\cis }{cis} \DeclareMathOperator {\Arg }{Arg} \DeclareMathOperator {\Rep }{Re} \DeclareMathOperator {\Imp }{Im} \DeclareMathOperator {\sech }{sech} \DeclareMathOperator {\csch }{csch} \DeclareMathOperator {\Log }{Log} \newcommand {\tightoverset }[2]{\mathop {#2}\limits ^{\vbox to -.5ex{\kern -0.75ex\hbox {$#1$}\vss }}} \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 {\utan }[0]{\vec {\hat {t}}} \newcommand {\unormal }[0]{\vec {\hat {n}}} \newcommand {\ubinormal }[0]{\vec {\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 {\vector }[1]{\left \langle #1\right \rangle }$

In this section we compute derivatives involving $a^x$ and $\log _a(x)$ .

The derivative of $a^x$

We begin by computing the derivative of exponential functions of the form $f(x) = a^x$ where $a > 0$ . To do this we will use the chain rule and a conversion formula to the natural base, $e$ . The conversion formula is $a^x = e^{x\ln (a)}.$

Now, by the chain rule,

$\dd {x} a^x = \dd {x} e^{x\ln (a)} = e^{x\ln (a)}\cdot \ln (a) = a^x \ln (a).$

This gives us the following theorem:

Let $a >1$ , then $\frac {d}{dx}\left (a^x\right ) = a^x \ln (a)$

If $a = e$ , then since $\ln (e) = 1$ , we have $\frac {d}{dx}\left (e^x\right ) = e^x \ln (e) =e^x$ which agrees with the formula we saw in section 2.3.

example 1 If $f(x) = 2^x$ then $f'(x) = 2^x \ln (2)$ and if $f(x) = 10^x$ then $f'(x) = 10^x \ln (10)$ .

(problem 1) Compute $\frac {d}{dx} \left (5^x\right ).$

Use the exponential rule with $a = 5$

The derivative of $a^x$ is $a^x\ln (a)$

The derivative of $5^x$ is $\answer {5^x\ln (5)}$ .

example 2 Find $f'(x)$ if $f(x) = 2^x + 3^x + 5^x.$
Using the Sum Rule we have, $f'(x) = 2^x\ln (2) + 3^x\ln (3) + 5^x\ln (5)$ .

(problem 2a) Compute $\frac {d}{dx} \left (10^x + 100^x\right )$

Use the Sum Rule

The derivative of $10^x + 100^x$ with respect to $x$ is $\answer {10^x \ln (10) + 100^x \ln (100)}$

(problem 2b) Compute $\frac {d}{dx} \left (e^x + \pi ^x\right )$

$\frac {d}{dx} \left (a^x\right ) = a^x \ln (a)$

The derivative of $e^x$ is itself

The derivative of $e^x + \pi ^x$ with respect to $x$ is $\answer {e^x + \pi ^x \ln (\pi )}$

(problem 2c) Compute $\frac {d}{dx} \left (3\cdot 2^x\right )$

Use the Constant Multiple Rule

The derivative of $3\cdot 2^x$ with respect to $x$ is $\answer {3\cdot 2^x \ln (2)}$

(problem 2d) Compute $\frac {d}{dx} \left [(x^4 + 5)2^x\right ]$

Use the Product Rule with $x^4 + 5$ and $2^x$ .

$(fg)' = f'g+fg'$ .

The derivative of $x^4 + 5$ is $4x^3$

The derivative of $2^x$ is $2^x \ln (2)$

The derivative of $(x^4 + 5)2^x$ with respect to $x$ is $\answer {4x^3 2^x + (x^4 + 5) 2^x \ln (2)}$

The Chain Rule applied to a general exponential function, $a^x$ , yields the following:

$\dd {x} \left [a^{g(x)}\right ] = a^{g(x)} \ln (a) \cdot g'(x).$

example 3 Find $y'$ if $y = 3^{\cos (x)}$
We use the chain rule with $\cos (x)$ as the inside function: $\dd {x} \left [3^{\cos (x)}\right ] = 3^{\cos (x)} \ln (3) \cdot (-\sin (x)) = -3^{\cos (x)} \ln (3) \sin (x).$

(problem 3) Compute $\dd {x} \left [ 4^{\sec (x)}\right ] = \answer {4^{\sec (x)}\ln (4) \sec (x)\tan (x)}.$

The derivative of $\log _a(x)$

To find the derivative of $\log _a(x)$ with $a > 0, a \neq 1$ , we use the following change of base formula: $\log _a(x) = \frac {\ln (x)}{\ln (a)}.$

Since the denominator in the above formula is a constant, we can use the Constant Multiple Rule to find the derivative: $\dd {x} \left [\log _a(x)\right ] = \dd {x} \left (\frac {\ln (x)}{\ln (a)}\right ) = \frac {1}{\ln (a)}\cdot \dd {x} \left [\ln (x)\right ] = \frac {1}{\ln (a)} \cdot \frac {1}{x} = \frac {1}{x\ln (a)}.$

This gives us the following theorem:

Let $a >1$ , then $\frac {d}{dx}\left (\log _a(x)\right ) = \frac {1}{x \ln (a)}$

If $a = e$ , then since $\ln (e) = 1$ , we have $\frac {d}{dx}\left (\ln x\right ) = \frac {1}{x \ln (e)} = \frac {1}{x}$ which agrees with the formula we saw in section 2.3.

example 4 The common log has base 10 and is usually written $\log (x)$ . So, if $f(x) = \log (x)$ then $f'(x) = \frac {1}{x\ln (10)}.$

(problem 4a) Compute $\ddx \left [\log _5(x)\right ] = \answer {\frac {1}{x\ln (5)}}.$

(problem 4b) Compute $\frac {d}{dx} \left [2\log (x)\right ]$

Common log: $\log (x) = \log _{10}(x)$

The derivative of $2\log (x)$ with respect to $x$ is $\answer {\frac {2}{x\ln (10)}}$

(problem 4c) Find $f'(1)$ if $f(x) = x^2 \log _2(x)$ .

$f'(1) = \answer {1/\ln (2)}.$

2024-09-27 13:55:07

1 The derivative of a^x

2 The derivative of \log _a(x)

The derivative of $a^x$

The derivative of $\log _a(x)$