3.5 Derivatives of exponential functions

Two of the functions are the same.

A-$(a)$ and $(d)$, B-$(e)$, C-$(c)$, D-$(b)$

Since $1^x=1$ for any $x$, we see that $(b)$ is just the constant function $y=1$, so D matches to $(b)$.

Since $2^{-x}=\frac{1}{2^x}=\left(\frac{1}{2}\right)^x$, functions $(a)$ and $(d)$ are the same. This is the only function out of the lot that grows as $x\to-\infty$ and shrinks as $x \to \infty$, so A matches to $(a)$ and $(d)$.

This leaves B and C to match to $(c)$ and $(e)$. Since $3>2$, when $x>0$, $3^x>2^x$. So, $(e)$ matches to the function that grows more quickly to the right of the $x$-axis: B matches to $(e)$, and C matches to $(c)$.

$(b),~(d),~(e)$

First, let's consider the behaviour of exponential functions $a^x$ based on whether $a$ is greater or less than 1. As we know, $\ds\lim_{x\to\infty}a^x=\left{\begin{array}{ll} \infty & a>1\0&a<1 \end{array}\right.$ and $\ds\lim_{x\to-\infty}a^x=\left{\begin{array}{ll} 0 & a>1\\infty&a<1 \end{array}\right.$. Our function has$\ds\lim_{x \to \infty} f(x)=\infty$and$\ds\lim_{x \to -\infty} f(x)=0$, so we conclude$a>1$: thus$(d)$and also$(b)$hold. (We could have also seen that$(b)$holds because$a^x$ is defined for all real numbers.)

It remains to decide whether $a$ is greater or less than $e$. (If $a$ were equal to $e$, then $f'(x)$ would be the same as $f(x)$.) We saw in the text that $\diff{}{x}\{a^x\}=C(a)a^x$ for the function $C(a)=\ds\lim_{h \to 0} \dfrac{a^h-1}{h}$. We know that $C(e)=1$. (Actually, we chose $e$ to be the number that has this property.) From our graph, we see that $f'(x)<f(x)$, so $C(a)<1=C(e)$. In other words, $\ds\lim_{h \to 0} \dfrac{a^h-1}{h}<\ds\lim_{h \to 0} \dfrac{e^h-1}{h}$; so, $a<e$. Thus $(e)$ holds.

When can you use the power rule?

False

The power rule tells us that $\diff{}{x}\{x^n\}=nx^{n-1}$. In this equation, the variable is the base, and the exponent is a constant. In the function $e^x$, it's reversed: the variable is the exponent, and the base it a constant. So, the power rule does not apply.

What is the shape of the curve $e^{ax}$, when $a$ is a positive consant?

increasing

$P(t)$ is an increasing function over its domain, so the population is increasing.

There are a few ways to see that $P(t)$ is increasing.

What we really care about is whether $e^{0.2t}$ is increasing or decreasing, since an increasing function multiplied by 100 is still an increasing function, and a decreasing function multiplied by 100 is still a decreasing function. Since $f(t)=e^t$ is an increasing function, we can use what we know about graphing functions to see that $f(0.2t)=e^{0.2t}$ is also increasing.

If you know the first derivative, this should be easy.

$e^x$

The derivative of $e^x$ is $e^x$: taking derivatives leaves the function unchanged, even if we do it 180 times. So $f^{(180)}=e^x$.

Simplify

(b) and (d)

We simplify the functions to get a better idea of what's going on.

($a$): $y=e^{3\log x}+1=\left(e^{\log x}\right)^3+1=x^3+1$. This is not a line. ($b$): $2y+5=e^{3+\log x}=e^3e^{\log x}=e^3x$. Since $e^3$ is a constant, $2y+5=e^3x$ is a line. ($c$): There isn't a fancy simplification here–this isn't a line. If that isn't a satisfactory answer, we can check: a line is a function with a constant slope. For our function, $y'=\diff{}{x}\{e^{2x} +4\}=\diff{}{x}\{e^{2x}\}=\diff{}{x}\left\{(e^x)^2\right\}=2e^xe^x=2e^{2x}$. Since the derivative isn't constant, the function isn't a line. ($d$): $y=e^{\log x}3^e+\log 2=3^e\,x+\log 2$. Since $3^e$ and $\log 2$ are constants, this is a line.