$$y=e^x$$​​

In a nutshell

A key property of exponential functions is that the graphs of their gradient functions are closely related to the graphs of the original function. In particular, the gradient function is just the original function multiplied by a constant. There exists a unique value for the base for an exponential function such that this constant is equal to $$1$$. This value is known as $$e$$​.

Taking the derivative of exponential functions

Let $$f(x) = a^x$$​, where $$a$$​ is a constant greater than $$0$$​. Then the gradient function, $$f'(x)$$​ can be shown to be equal to $$kf(x)$$, where $$k$$​ is a constant. The value of $$k$$​ depends on $$a$$, and can be potentially any real number depending on the choice of $$a$$​.

When $$a = 1$$​, the function $$f(x) = 1^x$$ is constant, so the gradient function is $$0$$, and hence $$k = 0$$​.

When $$a = 2$$​, the value of $$k$$​ turns out to be equal to $$0.693\dots$$​, so $$f'(x) = 0.693\dots \times f(x)$$. This means that the rate at which the function is growing at any point $$x$$​ is less than the value of the function evaluated at $$x$$, as $$k < 1$$​.

When $$a = 3$$​, the value of $$k$$​ turns out to be equal to $$1.099\dots$$​, so $$f'(x) = 1.099\dots \times f(x)$$​. This means that the rate at which the function is growing at any point $$x$$​ is greater than the value of the function evaluated at $$x$$, as $$k > 1$$​.

The constant $$e$$​​

In fact, there exists a unique value of $$a$$​ between $$2$$​ and $$3$$​ for which ​the corresponding value of $$k$$​ will be precisely $$1$$. The first few digits of this value work out to be $$2.71828\dots$$ and the exact value is referred to as $$e$$. This is one of the most important mathematical constants, and is ubiquitous in many mathematical subfields and mathematically adjacent disciplines, much like $$\pi$$​.

So, in particular, if $$a = e$$, then $$f(x) = e^x$$​ and $$f'(x) = 1 \times f(x) = e^x$$​. This means that the rate at which the function is growing at any point $$x$$​ is equal to the value of the function at $$x$$​.

In general, if you are given a function $$f(x)$$​ with gradient function $$f'(x)$$​, and another function $$g(x)$$​ such that $$g(x) = f(kx)$$​ for some constant $$k$$​, then the gradient function is given by $$g'(x) = kf'(kx)$$.

So, if you have that $$f(x) = e^{kx}$$​, then $$f'(x) = ke^{kx}$$​. If $$y = e^{kx}$$​, then $$\dfrac{dy}{dx} = ke^{kx}$$​.

Example 1

What is the gradient of the curve with equation $$y = e^{2-x}$$ at the point where $$x=3$$?

Observe that you can rewrite the equation to factor out a constant term:

$$\begin{aligned}y &= e^{2-x}\\&= e^2 \times e^{-x}\end{aligned}$$​​

The derivative of the function ​$$f(x) = e^{-x}$$ is computed, following the above discussion, to be $$f'(x) = -e^{-x}$$.

Therefore, get:

$$\begin{aligned}y &= e^{2-x}\\y &= e^2 \times e^{-x}\\\dfrac{dy}{dx} &= e^2 \times -e^{-x}\\\dfrac{dy}{dx} &= -e^{2-x}\end{aligned}$$​​

To find the gradient of the curve with equation $$y = e^{2-x}$$ at the point where $$x=3$$, substitute $$x=3$$ into $$\dfrac{dy}{dx}$$:

​$$\begin{aligned}\dfrac{dy}{dx} &= -e^{2-x}\\&= -e^{2-3}\\&= -e^{-1}\end{aligned}$$​​

Therefore, the gradient of the curve with equation $$y = e^{2-x}$$ at the point where $$x=3$$ is equal to $$\underline{-e^{-1}}$$.

Example 2

Sketch the graph of the equation $$y = 5e^{-2x}$$. Give the coordinates of any points where the graph crosses the axes.

Given an equation $$y = f(x)$$, the graph of the equation $$y = kf(x)$$, where $$k$$​ is a positive constant, is the graph of the equation $$y = f(x)$$​ scaled in the $$y$$-direction by a factor of $$k$$.

The graph of the equation $$y = f(kx)$$ is the graph of the equation $$y = f(x)$$ scaled in the $$x$$​-direction by a factor of $$\dfrac1k$$ .

The graph of the equation $$y = f(-x)$$ is the graph of the equation $$y = f(x)$$ reflected about the $$y$$-axis.

Putting all of these together, if $$f(x) = e^x$$, then $$5e^{-2x} = 5f(-2x)$$.

So, the graph of the equation $$y = 5e^{-2x}$$ is the graph of the equation $$y = e^x$$ scaled by a factor of $$5$$ in the $$y$$-direction, a factor of $$\dfrac12$$ in the $$x$$-direction, and reflected about the $$y$$-axis. The graph of $$y = e^x$$ only crosses either axis at the point $$(0, 1)$$, so it follows this graph only crosses either axis at the point $$\left(\dfrac02,5 \times 1\right) = (0,5)$$.

Therefore, the only point the graph of the equation $$y = 5e^{-2x}$$ crosses either axis is $$\underline{(0,5)}$$.

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