Revision:Exponentials and logarithms

The exponential function is the function whose derivative is equal to its equation. In other words:

\frac{\text{d}}{\text{d}x}e^x = e^x

Because of this special property, the exponential function is very important in mathematics and crops up frequently.


Note: You may choose to leave this portion unheeded, for knowledge of this may not be required by your syllabus.

We shall begin with the differentiation of a simple exponential function - y = a^x, w.r.t.  x - from the first principles. If you're unfamiliar with differentiation with first principle, might I recommend you first direct your notice yonder: Differentiation from First Principles.

For the points (x,y) the equation stands:

y = a^x

Hence for the points (x+\delta x,y+\delta y), where \delta x is a small change in x and \delta y the corresponding small change in y, the equation gives:

y+\delta y = a^{x+\delta x}

Substituting y = a^x into the equation

a^x+\delta y = a^{x+\delta x}

\delta y = a^{x+\delta x} - a^x

\delta y = a^xa^{\delta x} - a^x

\delta y = a^x(a^{\delta x} - 1)

Dividing both sides by \delta x gives:

\displaystyle \frac{\delta y}{\delta x}=a^x(\frac{a^{\delta x} - 1}{\delta x})

\displaystyle \lim_{\delta x \to 0} \frac{\delta y}{\delta x} = \frac{dy}{dx} = a^x \lim_{\delta x \to 0} \frac{a^{\delta x}-1}{\delta x}

Isolating the part: \displaystyle \frac {a^{\delta x} - 1}{\delta x} and further computation yields the culmination of this proof. Taking the value of a to be 3 and then applying values of x that approach zero, we can deduce that  \displaystyle \lim_{\delta x \to 0} \frac{3^{\delta x} - 1}{\delta x} \approx 1.09861

From this one can ascribe veracity to the relationship \displaystyle \lim_{\delta x \to 0} \frac {a^{\delta x} - 1}{\delta x}=k. Gleaning from it one can also deduce that the value of k depends upon the value of a and for the value of k to be equal to 1, the value of a must be smaller than 3.

To find the value of a, where k=1, we use the same relationship.

\displaystyle \frac {a^{\delta x} - 1}{\delta x} \approx 1

a^{\delta x} \approx 1+\delta x

a \approx (1+\delta x)^{\frac{1}{\delta x}}

Let \displaystyle \delta x = \frac{1}{n}, which by extension means \displaystyle \frac{1}{\delta x}=n and

a=\lim_{\delta x \to 0}(1+\delta x)^{\frac {1}{\delta x}}

Representing \delta x as an expression of n, gives us:

\displaystyle a=\lim_{\delta x \to \infty}(1+\frac{1}{n})^n

The value that is found using values of n, as it approaches infinity gives the us the following value for a:

a \approx 2.71828

This value of a, one of the most beautiful and prolific numbers in all of mathematics, is known as Euler's number or as you may know it e.

Putting the value of a equal to e, and thence you can derive:

\frac{\text{d}}{\text{d}x}e^x = e^x

Alternative Method

This method of proving the derivative of e^x is e^x, though simpler, requires one to draw from the chain rule and the differentiation of ln x. A revision and the proper derivation of the function ln x can be found below.

The Chain Rule:

Let there be an equation defined thus: y=(x+1)^2

Let u be equal to x + 1

And hence y=u^2

If such a case can be made the chain rule declares. (Note: these should not be thought of as fractions)

\displaystyle \frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dx}

In this case:

\displaystyle \frac{du}{dx}=\frac{d}{dx}(x+1)=1

\displaystyle \frac{dy}{du}=\frac{d}{du}(u^2)=2u

\displaystyle \frac{dy}{dx}=2u \times 1= 2(x+1)

The derivative of ln x is \frac{1}{x}

Let there be function defined as y=ln e^x

One of the logarithmic properties dictates that the power of the number of whose logarithm is to be find, can be brought down to be multiplied with the rest of the function.

Hence it follows that the above function can be written as:

y=x \times ln e

As ln e = 1 the function now becomes y=x. Hence the differential of this function is:


The differential can also be found using Chain Rule.

Let u=e^x

The function hence becomes y=lnu

Hence \displaystyle \frac{dy}{du}=\frac{d}{du}(lnu)=\frac{1}{u}=\frac{1}{e^x}

\displaystyle \frac{du}{dx}=\frac{d}{dx}(u)=\frac{d}{dx}(e^x)

\displaystyle \frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dx}=\frac{1}{e^x} \times \frac{d}{dx}(e^x)=1

Hence \displaystyle \frac{d}{dx}(e^x)=1 \times e^x

\displaystyle \frac{d}{dx}(e^x)=e^x


Like most functions, the exponential has an inverse function, which is \ln x (pronounced 'log x').


f(x) = e^x \Leftrightarrow f^{-1}(x) = ln x

The Natural Logarithm

\ln x is also known as the natural logarithm. The derivative of \ln x is \frac{1}{x}:

\displaystyle \frac{d (\ln x)}{dx} = \frac{1}{x}

It therefore follows that:

\displaystyle \int \frac{1}{x} = \ln x + c.


This article is not detailed enough for all you need to know for C3 modules.