Calculus of Exponential Functions

Lesson

If we try to find the gradient function from first principles for an exponential function $a^x$`a``x` where the base $a$`a` is an arbitrary positive number $\ne1$≠1, we quickly reach an impasse. For example, if $f(x)=a^x$`f`(`x`)=`a``x`, then

$f'(x)$f′(x) |
$=$= | $\lim_{h\rightarrow0}\frac{a^{x+h}-a^x}{h}$limh→0ax+h−axh |

$=$= | $a^x\left(\lim_{h\rightarrow0}\frac{(a^h-1}{h}\right)$ax(limh→0(ah−1h) |

and it is not obvious what the limit $\lim_{h\rightarrow0}\frac{a^h-1}{h}$lim`h`→0`a``h`−1`h` should be.

Try putting $a=2$`a`=2 and make $h=0.01$`h`=0.01 using your calculator. With these values of $a$`a` and $h$`h`, we find $\frac{a^h-1}{h}\approx0.69$`a``h`−1`h`≈0.69.

Now try $a=3$`a`=3 with $h=0.01$`h`=0.01. This gives $\frac{a^h-1}{h}\approx1.1$`a``h`−1`h`≈1.1.

We could try other values of $a$`a` and perhaps smaller values of $h$`h` to get better estimates of the limiting values as $h\rightarrow0$`h`→0. These numerical results suggest that the value of $\lim_{h\rightarrow0}\frac{a^h-1}{h}$lim`h`→0`a``h`−1`h` increases with increasing values of $a$`a`, and it is less than $1$1 when $a=2$`a`=2 but greater than $1$1 when $a=3$`a`=3.

Thus, we suspect that there is a value of $a$`a` somewhere between $2$2 and $3$3 such that $\lim_{h\rightarrow0}\frac{a^h-1}{h}=1$lim`h`→0`a``h`−1`h`=1.

Further experiment shows that this number is somewhere near $2.7$2.7. You should check that $\frac{2.7^{0.01}-1}{0.01}\approx0.998$2.70.01−10.01≈0.998.

This special base is given the symbol $e$`e`. It follows that the exponential function with this base, $f(x)=e^x$`f`(`x`)=`e``x`. has derivative given by

$f'(x)$f′(x) |
$=$= | $e^x\left(\lim_{h\rightarrow0}\frac{e^h-1}{h}\right)$ex(limh→0eh−1h) |

$=$= | $e^x$ex |

To differentiate a function $f(x)=ka^x$`f`(`x`)=`k``a``x` where $a>1$`a`>1 but $a\ne e$`a`≠`e`, we need to express the function using the base $e$`e`. This amounts to a change of the relative scales on the axes in a graphical representation and the kind of transformation involved was introduced elsewhere.

We require $y$`y` such that $a^x=e^y$`a``x`=`e``y`.

Taking the logarithm base $e$`e` (natural logarithm $\ln$`l``n`) of both sides, we have $\ln a^x=y$`l``n``a``x`=`y` and thus, $y=x\ln a.$`y`=`x``l``n``a`. Therefore, $a^x$`a``x` is equivalent to $e^{x\ln a}$`e``x``l``n``a`.

Therefore, if $f(x)=ka^x$`f`(`x`)=`k``a``x`, we have

$f(x)$f(x) |
$=$= | $ke^{x\ln a}$kexlna |

$\therefore\ \ f'(x)$∴ f′(x) |
$=$= | $k\ln a.e^{x\ln a}$klna.exlna |

$=$= | $k\ln a.a^x$klna.ax |

We used the rule for differentiating a function multiplied by a constant: $\frac{\mathrm{d}}{\mathrm{d}x}\left(k.f(x)\right)=k.f'(x)$`d``d``x`(`k`.`f`(`x`))=`k`.`f`′(`x`).

And we used the function-of-a-function rule applied to the exponential function: $\frac{\mathrm{d}}{\mathrm{d}x}\left(e^{mx}\right)=e^{mx}.m$`d``d``x`(`e``m``x`)=`e``m``x`.`m`.

And finally, we wrote $e^{x\ln a}$`e``x``l``n``a` in the equivalent form $a^x$`a``x`.

Differentiate $g(x)=1.5\times2^x$`g`(`x`)=1.5×2`x`.

We change the base to obtain the equivalent representation $g(x)=1.5\times e^{\ln2.x}$`g`(`x`)=1.5×`e``l``n`2.`x`. Then, $g'(x)=1.5\times\ln2.e^{\ln2.x}=1.5\times\ln2\times2^x$`g`′(`x`)=1.5×`l``n`2.`e``l``n`2.`x`=1.5×`l``n`2×2`x`.

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

Apply differentiation methods in solving problems