2.4: General Exponential and Logarithmic Functions

For a general exponential function $y = a^x$, with $a > 0$, use logarithmic differentiation to find its derivative:

$$\begin{aligned} \ln\,y ~&=~ \ln\,\left(a^x\right) ~=~ x \,\ln\,a\\ \ddx\,(\ln\,y) ~&=~ \ddx\,(x \,\ln\,a) ~=~ \ln\,a\\ \frac{y'}{y} ~&=~ \ln\,a \quad\Rightarrow\quad y' ~=~ y \cdot \ln\,a\end{aligned} \nonumber$$

Thus, the derivative of $y = a^x$ is: In general, for an exponent of the form $u = u(x)$:

Find the derivative of $y = 2^{\cos x}$.

Solution: This is the case where $a = 2$, so:

$$\dydx = (\ln\,2)\;2^{\cos x} \;\cdot\; \ddx\,(\cos x) ~=~ -(\ln\,2)\;(\sin x)\;2^{\cos x} \nonumber$$

Note that any exponential function $y = a^x$ can be expressed in terms of the exponential function $e^x$. Since

$$a^x ~>~ 0 \quad\Rightarrow\quad e^{\ln\,(a^x)} ~=~ a^x ~~, \nonumber$$

and since $\ln\,(a^x) = x\,\ln\,a$, then: Computers and calculators often use the above formula to calculate $a^x$.

The function $y = a^x$ has an inverse for any $a > 0$, except for $a = 1$ (in that case $y = 1^x = 1$ is just a constant function). To see this, notice that since $a^x > 0$ for all $x$, and $\ln\,a < 0$ for $0 < a < 1$, while $\ln\,a > 0$ for $a > 1$, then $\dydx = (\ln\,a)\,a^x$ is always negative if $0 < a < 1$ and always positive if $a > 1$. Thus, $y = a^x$ is a strictly decreasing function if $0 < a < 1$, and it is a strictly increasing function if $a > 1$. The graphs in each case are shown in Figure [fig:expa].

Hence, for any $a >0$ with $a \ne 1$ the function $f(x) = a^x$ is one-to-one, so it has an inverse function, called the base $\bm{a}$ logarithm and denoted by $f^{-1}(x) = \log_a x$. It is often spoken as “log base $a$ of $x$”. The graphs for $a < 1$ and $a > 1$ are shown in Figure [fig:loga]. Note that the natural logarithm is just the base $a$ logarithm in the special case with $a = e$, i.e. $\ln\,x = \log_e x$. The base $a$ logarithm has properties similar to those of the natural logarithm (and the corresponding properties of $a^x$):

Note that $\log_a x$ can be put in terms of the natural logarithm, since

$$x ~=~ a^{\log_a x} \quad\Rightarrow\quad \ln\,x ~=~ \ln\,\left(a^{\log_a x}\right) ~=~ (\log_a x) \cdot (\ln\,a) \nonumber$$

so dividing the last expression by $\ln\,a$ gives: The above formula is useful on calculators that do not have a $\log_a x$ key or function. Taking the derivative of both sides yields: In general, when taking the logarithm of a function $u = u(x)$:

Find the derivative of $y = \log_2 (\cos\,4x)$.

Solution: This is the case where $a = 2$, so:

$$\dydx = \frac{1}{(\cos\,4x)\,(\ln\,2)} \;\cdot\; \ddx\,(\cos\,4x) ~=~ -\frac{4 \sin\,4x}{(\ln\,2)\,(\cos\,4x)} \nonumber$$

The number $a$ is the base of both the logarithm function $\log_a x$ and the exponential function $a^x$. Base 2 and base 10 are the most commonly used bases other than base $e$. Base 10 is how numbers are normally expressed, as combinations of powers of 10 (e.g. $2014 = \bm{2} \cdot 10^3 \;+\; \bm{0} \cdot 10^2 \;+\; \bm{1} \cdot 10^1 \;+\; \bm{4} \cdot 10^0$). Base 2 is especially useful in computer science, since computers represent all numbers in binary format, i.e. as a sequence of zeros and ones, indicating how many successive powers of two to take and then sum up.⁸ For example, the number 6 is represented in binary format as 110, since $\bm{1} \cdot 2^2 \;+\; \bm{1} \cdot 2^1 \;+\; \bm{0} \cdot 2^0 ~=~ 4 + 2 + 0 = 6$.

[sec2dot4]

For Exercises 1-9, find the derivative of the given function.

3

$y = \dfrac{3^x ~+~ 3^{-x}}{2}$

$y = 2^{\ln\,3x}\vphantom{\dfrac{3^x}{2}}$

$y = 2^{2^x}\vphantom{\dfrac{3^x}{2}}$

3

$y = \tan^{-1} \pi^x$

$y = \log_2 \,(x^2 + 1)$

$y = \log_{10} \,e^x$

3

$y = \sin\,\left(\log_2 \,\pi x\right)$

$y = \log_2 \,4^{2x}$

$y = 8^{\log_2 \,x}$

[[1.]]

Show that for all constants $k$ the function $y = A a^{\frac{kx}{\ln\,a}}$ satisfies the differential equation $\dydx = ky$. Does this contradict the statement made in Section 2.3 that the only solution to that differential equation is of the form $y = A e^{kx}$? Explain your answer.