Having trouble with differentiation of logarithms? You’re not alone! In this article, we’ll start with the basics, and slowly work in differentiation rules to build up to more difficult questions. With our worked examples, you can build up your knowledge of Year 11 logarithms.
In this article, we will be utilising our knowledge of logarithm laws and the relationship between exponentials and logarithms to understand the derivative of logarithms. You’ll also need knowledge of basic calculus, including differentiation rules.
Logarithms are taught across Year 11 and Year 12. In the real world, exponentials and logarithms can be used to represent and manipulate multiplicative growth and decay rates. If you plan to pursue a career in a STEM field, you will definitely encounter logarithmic functions. In your HSC, it is quite common to see calculus questions involving logarithms.
We’re going to look at the following NESA syllabus points. These syllabus points are spread across the Year 11 and Year 12 Advanced Mathematics outcomes. Intuitively, these syllabus points can be grouped and taught together.
NESA requires students for Stage 6 Maths Advanced to demonstrate proficiency in the following dot points:
E1.3 The exponential function and natural logarithms
C2.1 Differentiation of trigonometric, exponential and logarithmic functions
C2.2 Rules of differentiation
Students should already be familiar with:
Natural logarithms and exponentials are intrinsically linked. We will use the relationship between them to find the derivative of a natural logarithm.
To differentiate \(y=lnx\):
Express the logarithm in its equivalent exponential form:
\(x=e^y\)Differentiate with respect to \(y\):
\(\frac{dx}{dy}=e^y\)
However, we are interested in \(\frac{dy}{dx}\):
\(\frac{dy}{dx}=\frac{1}{dx/dy}=\frac{1}{e^y}\)
Since we know \(x=e^y\):
\(\frac{dy}{dx}=\frac{1}{x}\)
Hence the derivative of a natural logarithm is:
\(\frac{d}{dx} [lnx]=\frac{1}{x}\)
It is not necessary to prove this result, but rather you should memorise it and know how to apply it to solve calculus question.
Most logarithms you come across will be natural logarithms with base \(e\) (\(e.g. log_e (x)\) ). However you may encounter logarithms with different bases. To differentiate \(y=log_a (x)\):
Use the Change of Base Rule:
\(y=log_a (x)=\frac{log_e (x)}{log_e (a)}=\frac{1}{log_e (a)} \times ln(x)\)
Since \(\frac{1}{log_e (a)}\) is a constant, and we know how to differentiate a natural logarithm:
\(\frac{dy}{dx} = \frac{1}{log_e (a)} \times \frac{1}{x}\)
Again, this proof is not examinable and this result can be applied as a formula:
\(\frac{d}{dx} [log_a (x)]=\frac{1}{ln(a)} \times \frac{1}{x}\)
Now that we know the derivative of a natural logarithm, we can apply existing Rules for Differentiation to solve advanced calculus problems. Consider the following examples:
To find the derivative of \(y=ln(x^5 )\), we can apply the Chain Rule.
\begin{align*} \frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dx}, \ where \ u=x^5 \\
=\frac{d}{du}ln(u) \times \frac{d}{dx} x^5 \\
=\frac{1}{u \times 5x^4} \\
=\frac{1}{x^5} \times 5x^4 \\
=\frac{5}{x} \end{align*}
To find the derivative of \(y= x^2 ln(x)\), we can apply the Product Rule.
\begin{align*} \frac{dy}{dx}=u \frac{dv}{dx} + v \frac{du}{dx}, \ where \ u=x^2 \ and \ v=ln(x) \\
=x^2 \times \frac{d}{dx} ln(x) + ln(x) \times \frac{d}{dx} x^2 \\
=x^2 \times \frac{1}{x} + ln(x) \times 2x \\
=x + 2xln(x) \end{align*}
To find the derivative of \(y=\frac{ln(x)}{x}\), we can apply the Quotient Rule.
\begin{align*} \frac{dy}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}{}}{v^2}, \ where \ u=ln(x) \ and \ v=x \\
=\frac{x\times\frac{d}{dx}ln(x)-ln(x)\times\frac{d}{dx}x}{x^2} \\
=x \times \frac{{1}{x} – ln(x) \times 1}{x^2} \\
=\frac{1-ln(x)}{x^2}
\end{align*}
As you can see, all of our normal Rules for Differentiation apply to logarithms too, however some questions can become quite complicated with many terms! As a general rule, we can try to use Logarithmic Laws to simplify our functions before differentiating.
Consider \(y=ln(x^5)\) from the above example on Chain Rule.
Hence, \begin{align*} y=ln(x^5 )=5 ln(x), \\
\frac{dy}{dx}=5 \times \frac{1}{x}=\frac{5}{x} \end{align*}
As you can see, we have obtained the same solution with much less work. Clever applications of log laws will make calculus involving logarithms much simpler.
Note: In Application of Differentiation problems or Graphing problems involving calculus, be careful of how log laws affect the domain of the function.
Understanding how to differentiate logarithms allows us to solve a variety of problems involving calculus and logarithmic function. Consider the following example:
The function \(y=\frac{ln(kx)}{x}\) has a gradient of \(3\) at \(x=e\). Find the value of the constant \(k\).
Since we are given a gradient, we should form an equation using the first derivative of the function.
Before differentiating, check if the function can be simplified using log laws.
\(y=\frac{ln(kx)}{x}=\frac{ln(k)+ln(x)}{x}=\frac{ln(k)}{x}+\frac{ln(x)}{x}=ln(k) \times x^{-1}+\frac{ln(x)}{x}\)Now we can differentiate.
\(\frac{dy}{dx}=ln(k) \times x^{-2} \times (-1)+\frac{1/x \times x -1 \times ln(x)}{x^2} \)
\(=-\frac{ln(k)}{x^2} + \frac{1-ln(x)}{x^2} \)
At \(x=e\), the gradient is \(3\), so we can form the equation:
\(3= \frac{-ln(k)}{e^2} +\frac{1-ln(e)}{e^2}\)
After simplifying the equation:
\(3= -\frac{ln(k)}{e^2} + \frac{1-1}{e^2}\)
\(3= -\frac{ln(k)}{e^2} +0\)
\(-3e^2=ln(k)\)
We can use the relationship between natural logarithms and exponentials to solve for k:
\(k=e^{-3e^2}\)Our Year 11 Maths Advanced courses are the expert guided solution to your Maths problems. Learn more now!
Now it’s time for you to check your understanding!
Have a go at these questions and compare your answers to our solutions.
1. Find the derivative of \(y=(ln x)^2\)
2. Find the derivative of \(y= log_3 (x^2-2x)\)
3. Find the derivative of \(y=(x^2-9)ln(x+3)\)
4. Given the function \(y=\frac{3x}{ln(x)}\), find its gradient at \(x=e\).
1. \(\frac{dy}{dx}=\frac{2ln(x)}{x}\)
2. \(\frac{dy}{dx}=\frac{1}{ln(3)} \times \frac{2x-2}{x^2-2x}\)
3. \(\frac{dy}{dx}=2xln(x+3)+x-3\)
4. \( Gradient=3 – \frac{3}{e}\)
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