# Differentiating e to the kx

Differentiating e to the x does not change the function, i.e. y and dy/dx are the same expression. The graph below 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.

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:

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

Be careful when differentiating multiples of these functions. The power and multiple must be multiples together. See the Examples.

## Examples of Differentiating e to the kx

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

find $h'(x)$.

We can write $h(x)$ 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:

$h'(x)=-2\times\frac{1}{4}e^{-2x}+3\times 5e^{3x}=-\frac{1}{2}e^{-2x}+15e^{5x}$.

A scientist drops a heated metal ball into a cooler liquid. After time $t$ seconds, the temperature, $T^\circ$, of the ball is

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

Find the temperature of the ball at the instant the scientist drops the ball 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 $377^\circ$. The rate of cooling is given by the derivative of $T$:

$T'(t)=-0.1\times 350e^{-0.1t}=-35e^{-0.1t}$.

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

$T'(10)=-35e^{-0.1\times 10}=-35e^{-1}=-12.88$

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

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