An Introduction to Calculus – Differentiation Properties and Rules – A Guide to Mathematical Studies (Archive)

Welcome back to our ongoing exploration of the captivating world of calculus! In our previous article, we embarked on a journey through the essential concepts of limits and differentiation, laying the foundation for a deeper understanding of this remarkable branch of mathematics. Today, we are delving even further into the heart of calculus by focusing on the intricacies of differentiation.

As we continue our calculus series, we’ll dive into the invaluable tools and techniques that make calculus a powerhouse for analyzing change and motion. Specifically, in this installment, we will unravel some of the properties and rules of differentiation.

In the pages that follow, we will explore the key principles of differentiation, including the power rule, product rule, quotient rule, and chain rule. These rules are the building blocks that enable us to dissect and understand the behavior of functions with precision. Along the way, we’ll demystify the magic behind these rules, providing you with a clear and intuitive grasp of their inner workings. Without further ado, let’s continue!

Taking this opportunity, I would also like to introduce several common derivatives of functions :

$\frac{d}{dx}\ln(x) =\frac1x$

$\frac{d}{dx} \sin(x) = \cos(x)$

$\frac{d}{dx}\cos(x)=-\sin(x)$

$\frac{d}{dx}e^{x} = e^{x}$

These can all be proven by first principles of differentiation, and is thus left as an exercise to the readers. We will demonstrate the proof of the differential of $\sin(x)$ , which is as follows.

Generated by ChatGPT.¹

Firstly, we need to define some of the key properties of the differential operator. The first of them is that it satisfies the addition rule – that is to say for any function with form $f(x)=a + b +\text{...}+z$ , where all terms are functions of $x$ , $\frac{d}{dx}f(x)=\frac{d}{dx}a+\frac{d}{dx}b+\text{...}+\frac{d}{dx}z$ . This is proven easily with the first principles of differentiation, and is hence left as a practice for the readers. Another notable rule is the power rule :

For a function of the form $f(x)=ax^n$ , the derivative of the function would be $\frac{d}{dx}f(x)=a(n)x^{(n-1)}$ .

To prove this, we utilize once again the first principle of differentiation and the binomial expansion – for those unfamiliar, it is defined as follows :

where $\begin{pmatrix}n\\r\end{pmatrix} = \frac{n!}{r!(n-r)!}$ .

First, we plug $f(x)=x^n$ into the definition of the derivative and use the Binomial Theorem to expand out the first term.

$f'(x)=\lim_{\Delta{x}\rightarrow0}\frac{(x+\Delta{x})^n-x^n}{\Delta{x}}=\lim_{\Delta{x}\rightarrow0}\frac{(x^n+n\Delta{x}\cdot{x^{n-1}}+...+\Delta{x}^n)-x^n}{\Delta{x}}$

Now since $x^n$ cancels, all of the terms have a factor of $\Delta{x}$ which again cancel with the denominator. This gives us $\lim_{\Delta{x}\rightarrow0}nx^{n-1}+\text{...}+\Delta{x}^{n-1}$ , which reduces to $nx^{n-1}$ as all the other terms go to 0. As we can easily see that the constant term $a$ can be factored out of the limit, the resulting derivative would be $\frac{d}{dx}f(x)=a(n)x^{(n-1)}$ .

Another rule is the product rule, which governs the derivative of two products. In mathematical notation it is as follows :

For a function with form $f(x)=uv$ , where $u$ and $v$ are functions of $x$ , the derivative of the function can be expressed as $\frac{d}{dx}uv=u(\frac{d}{dx}v)+v(\frac{d}{dx}u)$ . This can be proven – you guessed it – by the first principle of differentiation. Below is an outline of the proof of the rule (an alternate method utilising logarithms can be used however this requires knowledge of implicit differentiation).

$\frac{d}{dx}f(x)g(x)=\lim_{\Delta{x}\rightarrow0}\frac{f(x+\Delta{x})g(x+\Delta{x})-f(x)g(x)}{\Delta{x}}$

$=\lim_{\Delta{x}\rightarrow0}\frac{f(x+\Delta{x})g(x+\Delta{x})+f(x+\Delta{x})g(x)-f(x+\Delta{x})g(x)-f(x)g(x)}{\Delta{x}}$

$=\lim_{\Delta{x}\rightarrow0}\frac{f(x+\Delta{x})(g(x-\Delta{x})+g(x))+g(x)(f(x+\Delta{x})-f(x))}{\Delta{x}}$

$=\lim_{\Delta{x}\rightarrow0}f(x+\Delta{x})\frac{g(x+\Delta{x})-g(x)}{\Delta{x}}+g(x)\lim_{\Delta{x}\rightarrow0}\frac{f(x+\Delta{x})-f(x)}{\Delta{x}}$

Now based on the definition of differentiation $\lim_{\Delta{x}\rightarrow0}f(x+\Delta{x})$ must be equal to $f(x)$ as a differentiable region must be continuous at all points. Applying the first principle of differentiation, we can easily see that $\lim_{\Delta{x}\rightarrow0}\frac{g(x+\Delta{x})-g(x)}{\Delta{x}}$ and $\lim_{\Delta{x}\rightarrow0}\frac{f(x+\Delta{x})-f(x)}{\Delta{x}}$ is the derivative of $f(x)$ and $g(x)$ respectively, hence we have successfully proven that $\frac{d}{dx}uv=u(\frac{d}{dx}v)+v(\frac{d}{dx}u)$ . A special case of this is the quotient rule, which states $\frac{d}{dx}\frac{u}{v}=\frac{v(\frac{d}{dx}u)-u(\frac{d}{dx}v)}{v^2}$ .

In fact, using this, we can now find the derivative of $\tan(x)$ . Using the identity $tan(x) = \frac{\sin(x)}{\cos(x)}$ and the quotient rule, we find that the derivative is $\frac{\cos(x)\cdot{\cos(x)}-[\sin(x)\cdot{(-\sin(x))}]}{\cos^2(x)}=\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}=\frac1{\cos^2(x)}$ (recall that $\sin^2(x)+\cos^2(x)=1$ ).

Now equipped with this knowledge, we may tackle the golden rule of differentiation – chain rule. This states that the derivative of $f(g(x))$ is $f^{\prime}(g(x))\cdot{g^\prime(x)}$ . This appears in many derivatives, for example $\sin(\ln(5x^2-2x))$ , where the derivative is obtained with three uses of chain rule, i.e. $\cos(\ln(5x^2-2x))\cdot\frac1{5x^2-2x}\cdot(10x-2)=\frac{2(5x-1)}{x(5x-2)}\cos(\ln(5x^2-2x))$ .

To prove this, we use first principles to take the derivative of $f(u(x))$ , where $f(x)$ and $u(x)$ are differentiable functions.

Generated by ChatGPT.

Practice Questions – Differentiate the following functions:

$f(x) = 3x^5 - 2x^3 + 7x$
$g(x) = x^2 \cdot \sin(x)$
$h(x) = \frac{x^2 - 3x + 2}{2x + 1}$
$k(x) = \sin(3x^2 + 2x)$

Now, equiped with this knowledge, differentiations of elementary functions should no longer pose any challenge. However, there are still functions that cannot be solved with these methods, for example $y = \arctan(x)$ . There are no known rules to us to take the differential directly – which means we need to introduce a new method of taking differentials of equations, named ‘Implicit Differentiation’. This is used when it is impossible or inconvenient to compute the derivative of $x$ explicitly for $y$ , as the name suggests. To do this, we need to treat $y$ as an implicit function of $x$ , such that when differentiating with respect to $x$ , $y$ becomes $\frac{dy}{dx}$ .

To differentiate $y = \arctan(x)$ , we first rearrange the equation into the form $\tan(y) = x$ . Then, by taking differential with respect to $x$ :

$=\frac1{\cos^2(y)}{\frac{dy}{dx}}=1$

$=\sec^2(x){\frac{dy}{dx}}=1$

$={\frac{dy}{dx}}=\frac1{\sec^2(y)}$ (Recall $\frac1{\cos(x)}=\sec(x)$ )

$={\frac{dy}{dx}}=\frac1{1-\tan^2(y)}$ (Use identity $1-\tan^2(x)=\sec^2(x)$ )

$={\frac{dy}{dx}}=\frac1{1-x^2}$ (Subsitute $y = \arctan(x)$ )

Thus using this we have successfully found the derivative of a non-elementary function. This, when used with conjunction with the above rules, means that most functions can now be differentiated with ease!

Practice Problems :

1. Find $\frac{dy}{dx}$ when $x^2 + y^2 = 25$ .

2. Differentiate $e^x + xy = 4$ with respect to $x$ .

3. Use implicit differentiation to find $\frac{dy}{dx}$ for the equation $x^3 + y^3 = 6xy$ .

In this journey through the realm of calculus, we’ve embarked on a quest to understand the fundamental principles of differentiation, exploring the intricacies of differentiation properties and rules. We’ve ventured into the world of limits, harnessed the power of the chain rule, tackled problems with the quotient rule, and navigated the intricacies of the product rule. Through this exploration, we’ve gained the tools to decipher the rate of change in any function and dissect complex equations with confidence.

As we conclude this introduction to differentiation, it’s crucial to remember that calculus is not just a mathematical discipline but a language of nature itself. It allows us to decipher the intricate patterns of the world around us, from the motion of celestial bodies to the behavior of atoms. It serves as the backbone of countless scientific and engineering breakthroughs, shaping the very foundations of modern technology.

As we draw this chapter on differentiation to a close, we stand at a crossroads in our exploration of calculus. While differentiation has enabled us to grasp the nuances of change, it’s equally essential to comprehend the cumulative effects and the grander picture. That’s where integration steps in. Just as we’ve dissected functions and unraveled their rates of change, integration allows us to piece together the fragments and uncover the accumulated quantities. It’s a natural progression in our journey through calculus, one that promises to reveal even deeper insights into the world of mathematics and its real-world applications. So, I shall see you next time with an introduction to integration – if you liked this writing, consider signing up below for my newsletter to get notified everytime I post a new piece of writing!

It took many iterations for ChatGPT to generate what was actually correct – do double-check its output before using it for any mathematical work. ↩︎

An Introduction to Calculus – Differentiation Properties and Rules

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