Interactive practice questions
We wish to determine the gradient function of $f\left(x\right)=e^x$f(x)=ex by first principles.
Using first principles, show that $f'\left(x\right)=e^x\lim_{h\to0}\left(\frac{e^h-1}{h}\right)$f′(x)=exlimh→0(eh−1h).
Complete the following table of values to determine the value of $\lim_{h\to0}\left(\frac{e^h-1}{h}\right)$limh→0(eh−1h). 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 |
| $\frac{e^h-1}{h}$eh−1h | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ | $\editable{}$ |
Hence what is the limiting value of $\frac{e^h-1}{h}$eh−1h as $h$h approaches $0$0?
Hence determine $f'\left(x\right)$f′(x).
Determine $f'\left(3.5\right)$f′(3.5).
Give your answer correct to two decimal places.
Which of the following is $f'\left(3.5\right)$f′(3.5) equal to?
$e^{2.5}$e2.5
$f\left(3.5\right)$f(3.5)
$3.5f\left(3.5\right)$3.5f(3.5)
Approximate $\frac{a^h-1}{h}$ah−1h for different values of $a$a by filling in the gaps in the table below. Round your answers to six decimal places where necessary.
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!+. . .
Differentiate $y=-4e^x$y=−4ex.