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2.05 Differentiation and exponential functions

Interactive practice questions

We wish to determine the gradient function of $f\left(x\right)=e^x$f(x)=ex by first principles.

a

Using first principles, show that $f'\left(x\right)=e^x\lim_{h\to0}\left(\frac{e^h-1}{h}\right)$f(x)=exlimh0(eh1h).

b

Complete the following table of values to determine the value of $\lim_{h\to0}\left(\frac{e^h-1}{h}\right)$limh0(eh1h). Write all values to four decimal places.

$h$h $-0.1$0.1 $-0.01$0.01 $-0.001$0.001 $0.001$0.001 $0.01$0.01 $0.1$0.1
$\frac{e^h-1}{h}$eh1h $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$
c

Hence what is the limiting value of $\frac{e^h-1}{h}$eh1h as $h$h approaches $0$0?

d

Hence determine $f'\left(x\right)$f(x).

e

Determine $f'\left(3.5\right)$f(3.5).

Give your answer correct to two decimal places.

f

Which of the following is $f'\left(3.5\right)$f(3.5) equal to?

$e^{2.5}$e2.5

A

$f\left(3.5\right)$f(3.5)

B

$3.5f\left(3.5\right)$3.5f(3.5)

C
Medium
9min

Approximate $\frac{a^h-1}{h}$ah1h for different values of $a$a by filling in the gaps in the table below. Round your answers to six decimal places where necessary.

Easy
1min

The value of $e^x$ex, where $e$e is the natural base, can be given by the expansion below:

$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\text{. . .}$ex=1+x1!+x22!+x33!+x44!+x55!+. . .

Medium
2min

Differentiate $y=-4e^x$y=4ex.

Easy
< 1min
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Outcomes

3.2.1.1

estimate the limit of 𝑎^(ℎ−1)/ℎ as ℎ→0 using technology, for various values of 𝑎>0

3.2.1.2

recognise that 𝑒 is the unique number 𝑎 for which the limit (in 3.2.1.1) is 1

3.2.1.4

establish and use the formula 𝑑/𝑑x (𝑒^𝑥)=𝑒^𝑥 and 𝑑/𝑑x 𝑒^[𝑓(𝑥)] = 𝑓′(𝑥)𝑒^𝑓(𝑥)

3.2.4.1

select and apply the product rule, quotient rule and chain rule to differentiate functions; express derivatives in simplest and factorised form

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