The diagram shows the graph of $y=e^{x}$ where e, sometimes known as Euler’s number, is given by e=2.718281828459… See more on this type of graph. The number e is special because everywhere on this graph, the gradient is the same as the y-coordinate.

It can be shown that the derivative of $y=e^{kx}$, where k is a constant, is $\frac{dy}{dx}=ke^{kx}$, i.e.:

$y=e^{kx},\hspace{10pt}\frac{dy}{dx}=ke^{kx}$

For k=1, this says that at each point on the graph of y, the gradient matches that of the y coordinate:

$y=e^{x},\hspace{10pt}\frac{dy}{dx}=e^{x}$

See more on differentiation.

## Example 1

Differentiate $y=2e^{4x}$.

$\frac{dy}{dx}=4\times 2e^{4x}=8e^{4x}$

## Example 2

Given that $h(x)=\frac{1+20e^{5x}}{4e^{2x}}$,
find h'(x).

h(x) can be written as

$h(x)=\frac{1}{4e^{2x}}+\frac{20e^{5x}}{4e^{2x}}=\frac{1}{4}e^{-2x}+5e^{3x}.$

We can differentiate more simply when h is written in this format:

$h$

## Example 3

After time t seconds, the temperature, T degrees, of a heated metal ball that is dropped into a liquid is given by

$T(t)=350e^{-0.1t}+27$

Find the temperature of the ball at the instant it is dropped into the liquid. Find the rate at which the ball is cooling after 10 seconds.

Substituting t=0 into T gives the intial temperature as 377 degrees. The rate of cooling is given by the derivative of T:

$T$

At time t=10 seconds, the rate of cooling is

$T$

to 2 decimal places. This means that the ball is cooling by 12.88 degrees per second but this is only true at the instant 10 seconds after the ball is dropped into the liquid.

See more on Growth & Decay.

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