New Zealand
Level 8 - NCEA Level 3

# Derivative of ka^x

Lesson

If we try to find the gradient function from first principles for an exponential function $a^x$ax 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)=ax, 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 $\lim_{h\rightarrow0}\frac{a^h-1}{h}$limh0ah1h should be.

### Experiment

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 $\frac{a^h-1}{h}\approx0.69$ah1h0.69.

Now try $a=3$a=3 with $h=0.01$h=0.01. This gives $\frac{a^h-1}{h}\approx1.1$ah1h1.1.

We could try other values of $a$a and perhaps smaller values of $h$h to get better estimates of the limiting values as $h\rightarrow0$h0. These numerical results suggest that the value of $\lim_{h\rightarrow0}\frac{a^h-1}{h}$limh0ah1h 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 $\lim_{h\rightarrow0}\frac{a^h-1}{h}=1$limh0ah1h=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$2.70.0110.010.998.

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

### other bases

To differentiate a function $f(x)=ka^x$f(x)=kax where $a>1$a>1 but $a\ne e$ae, 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 $a^x=e^y$ax=ey.

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

Therefore, if $f(x)=ka^x$f(x)=kax, 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: $\frac{\mathrm{d}}{\mathrm{d}x}\left(k.f(x)\right)=k.f'(x)$ddx(k.f(x))=k.f(x).

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

And finally, we wrote $e^{x\ln a}$exlna in the equivalent form $a^x$ax.

#### Example

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

We change the base to obtain the equivalent representation $g(x)=1.5\times e^{\ln2.x}$g(x)=1.5×eln2.x. Then, $g'(x)=1.5\times\ln2.e^{\ln2.x}=1.5\times\ln2\times2^x$g(x)=1.5×ln2.eln2.x=1.5×ln2×2x.

### Outcomes

#### M8-11

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

#### 91578

Apply differentiation methods in solving problems