A Brief Introduction to Calculus: Part 1.1 - Basic Differentiation Rules

  • Math
  • Calculus
  • Further Math
Published on Sun Jul 31 2022. Views
An introduction to basic concepts in Calculus; mostly based on the IGCSE Further Math textbook.

Introduction

This series would primarily focus on the three fundamental areas of calculus - derivatives, limits, and integration, looking briefly at some of the different concepts to get started with calculus.; thus no prior knowledge of calculus is needed.

This is the beginning of the first part of the series — Derivatives.

Note: the content covered here would only require a solid understanding of basic algebra.

So now, let's sit back, relax, and have some fun!

Quick Access

Functions

This is one of the most fundamental and critical ideas in calculus, so do ensure that you understand this part perfectly.

Functions are the way we define the relationship between an independent variable and a dependent variable. Like in computer science, a function has an input (mathematically called a variable) and an output (the value of a function).

For example, a quadratic equation in the form of y=ax2+bx+cy = ax^2 + bx + c could be written as f(x)=ax2+bx+cf(x) = ax^2 + bx + c where y=f(x)y = f(x).

In the example above, ax2+bx+cax^2 + bx + c would be the expression for the value of the function f(x)f(x), i.e. the output, but it's also a representation of the relationship between f(x)f(x), or yy, and xx; likewise xx would be the variable, i.e. the input.

Graphically, this is the equivalent of for each x-coordinate, calculate its corresponding y-coordinate using f(x)f(x); and the point P(x,f(x))P(x, f(x)) would hence be plotted, for example, on a cartesian graph.

Derivatives

Differentiation in calculus refers to the process of finding the derivative of a function, which also represents the rate of change of a function.

The way I think of taking a derivative is a linear downward transformation since the indices of x are reduced by 1.

The mathematical representation of the derivative of a function is f(x)f'(x) where the number of apostrophes after ff indicates the nth derivative of function f(x)f(x); meanwhile, you can also say that the first derivative of yy, with respect to xx, is dydx\frac{dy}{dx}. Furthermore, the nthnth derivative of yy is written as dnydxn\large\frac{d^{n}y}{dx^{n}}

Not only can we calculate the derivative of a function, but we can also calculate the derivative of a derivative. This might sound a bit confusing, but to put it simply, say we have a function f(x)f(x), its first derivative is f(x)f'(x), and the derivative of the first derivative, i.e. the second derivative of f(x)f(x), would be f(x)f''(x), so on and so forth.

A function's derivative can be calculated by applying some rules to each term of the expression.

Formulae for Differentiation

Here we will just discuss a few basic ones used in differentiation.

xnx^n

ddx(xn)=nxn1\frac{d}{dx}(x^n) = n \cdot x^{n - 1}

In other words, move the previous power of x out for multiplication and subtract 1 to get the new index.

Examples:

  • ddx(x3)=3x2\frac{d}{dx}(x^3) = 3x^2
  • ddx(x1)=x2\frac{d}{dx}(x^{-1}) = -x^{-2}
  • ddx(5x2)=52x=10x\frac{d}{dx}(5x^2) = 5 * 2x = 10x, note: this can be proved with the product rule, which would be discussed later

A Single Constant

When differentiating a single constant (a term that does not involve the variable that we are differentiating with respect to), like 7, π\pi, etc., simply ignore the term in the resulting derivative.

To understand why, though, we can let cc be a constant, then it is also the equivalent of cx0c \cdot x^{0} and the value is unchanged as anything that 0\not= 0 to the power of 0 is 1.

Therefore, applying the rule above would give 0cx10 \cdot c \cdot x^{-1}, which would always equal 0.

Examples:

  • ddx(4)=0\frac{d}{dx}(4) = 0
  • ddx(π)=0\frac{d}{dx}(\pi) = 0
  • ddx(t2)=0\frac{d}{dx}(t^2) = 0 note that ddx\frac{d}{dx} signifies that we should any variable that's not x should be treated as a constant, thus the derivative of the expression with respect to x is 0.
  • ddx(3px4)=12px3\frac{d}{dx}(3px^4) = 12px^3 in this case, p is treated as a constant that multiplies with 3 to make the coefficient of x4x^4.
  • ddx(3x210x+5)=6x10\frac{d}{dx}(3x^2 - 10x + 5) = 6x - 10

Sine & Cosine

ddx sin(ax)=acos(ax)ddx cos(ax)=asin(ax)\begin{align*} \frac{d}{dx}\space& \sin(ax) = a \cdot \cos(ax) \\ \\ \frac{d}{dx}\space& \cos(ax) = -a \cdot \sin(ax) \end{align*}

Note that it's easier to have your angle(if in degrees) converted to radians before differentiating. (Hint: π\pi radians = 180180^\circ)

Examples:

  • ddx[sin(3x)]=3cos(3x)\frac{d}{dx}[\sin(3x)] = 3\cos(3x)
  • ddx[12sin(4x)]=2cos(4x)\frac{d}{dx}[\frac{1}{2}\sin(4x)] = 2\cos(4x)
  • ddx[2cos(4x)]=8sin(4x)\frac{d}{dx}[2\cos(4x)] = -8\sin(4x)

The Exponential Function

ddx eax=aeax\frac{d}{dx}\space e^{ax} = a \cdot e^{ax}

Examples:

  • ddx ex=ex\frac{d}{dx}\space e^{x} = e^{x}
  • ddx 3e4x=12e4x\frac{d}{dx}\space 3e^{4x} = 12e^{4x}

Logarithmic Functions

ddx[ln(u)]=uu\frac{d}{dx}[\ln(u)] = \frac{u'}{u}

Where uu' is the derivative of uu.

Examples:

  • ddx ln(xn)=nx\frac{d}{dx}\space \ln(x^n) = \frac{n}{x}
  • ddx ln(sin(x))=cos(x)sin(x)=1tan(x)=cot(x)\frac{d}{dx}\space \ln(\sin(x)) = \frac{\cos(x)}{\sin(x)} = \frac{1}{\tan(x)} = \cot(x)
  • ddx ln(cos(x))=sin(x)cos(x)=tan(x)\frac{d}{dx}\space \ln(\cos(x)) = -\frac{\sin(x)}{\cos(x)} = -\tan(x)
  • ddx ln(c)=0\frac{d}{dx}\space \ln(c) = 0
  • ddx ln(c)=0\frac{d}{dx}\space \ln(c) = 0
  • ddx ln(eax)=a\frac{d}{dx}\space \ln(e^{ax}) = a

Rules of Differentiation

The Chain Rule

This is when an expression can be seen as a function nested inside another.

If y=f(g(x))y = f(g(x)), then yy' or dydx=f(g(x))g(x)\frac{dy}{dx} = f'(g(x)) \cdot g'(x).

Examples:

  1. Differentiate sin(2x2+3)\sin(2x^2 + 3)
f(x)=sin(x)f(x)=cos(x)g(x)=2x2+3g(x)=4xdydx=cos(2x2+3)4x\begin{align*} f(x) &= \sin(x) \\ f'(x) &= \cos(x) \\ \\ g(x) &= 2x^2 + 3 \\ g'(x) &= 4x \\ \\ \frac{dy}{dx} &= \cos(2x^2 + 3) \cdot 4x \end{align*}
  1. Differentiate ex2+x\large e^{x^2+x}
f(x)=exf(x)=exg(x)=x2+xg(x)=2x+1dydx=ex2+x(2x+1)\begin{align*} f(x) &= e^x \\ f'(x) &= e^x \\ \\ g(x) &= x^2 + x \\ g'(x) &= 2x + 1 \\ \\ \frac{dy}{dx} &= e^{x^2 + x}(2x + 1) \end{align*}
  1. Differentiate (x3+4)6(x^3 + 4)^6
f(x)=x6f(x)=6x5g(x)=x3+4g(x)=3x2dydx=6(x3+4)53x2=18x2(x3+4)5\begin{align*} f(x) &= x^6 \\ f'(x) &= 6x^5 \\ \\ g(x) &= x^3 + 4 \\ g'(x) &= 3x^2 \\ \\ \frac{dy}{dx} &= 6(x^3 + 4)^5 \cdot 3x^2 \\ &= 18x^2(x^3 + 4)^5 \end{align*}

The Product Rule

This is used to differentiate a term that is made by multiplying two others, e.g. sin(x)x3\sin(x)\cdot x^3

If y=uvy = uv, then dydx=uv+vu\frac{dy}{dx} = u \cdot v' + v \cdot u'.

Examples:

  1. Differentiate y=exsin(x)y = e^x \sin(x)
u=exu=exv=sin(x)v=cos(x)dydx=excos(x)+exsin(x)=ex[cos(x)+sin(x)]\begin{align*} u &= e^x \\ u' &= e^x \\ \\ v &= \sin(x) \\ v' &= \cos(x) \\ \\ \frac{dy}{dx} &= e^x\cos(x) + e^x\sin(x) \\ &= e^x[\cos(x) + \sin(x)] \end{align*}
  1. Differentiate y=x2(3+2x)4y = x^2(3 + 2x)^4
u=x2u=2xv=(3+2x)4v=8(3+2x)3dydx=8x2(3+2x)3+2x(3+2x)4=2x(3+2x)3[4x+(3+2x)]=2x(3+2x)3[6x+3]=6x(3+2x)3[2x+1]\begin{align*} u &= x^2 \\ u' &= 2x \\ \\ v &= (3 + 2x)^4 \\ v' &= 8(3 + 2x)^3\\ \\ \frac{dy}{dx} &= 8x^2(3+2x)^3 + 2x(3+2x)^4 \\ &= 2x(3+2x)^3[4x + (3 + 2x)] \\ &= 2x(3+2x)^3[6x + 3] \\ &= 6x(3+2x)^3[2x + 1] \end{align*}

The Quotient Rule

To differentiate a quotient (an expression obtained from dividing one value by another).

If y=uvy = \frac{u}{v}, then dydx=vuuvv2\frac{dy}{dx} = \large\frac{vu' - uv'}{v^2}.

Examples:

  1. Differentiate y=x3x+4y = \frac{x}{3x + 4}
u=xu=1v=3x+4v=3dydx=3x+43x(3x+4)2=4(3x+4)2\begin{align*} u &= x \\ u' &= 1 \\ \\ v &= 3x + 4 \\ v' &= 3 \\ \\ \frac{dy}{dx} &= \frac{3x+4 - 3x}{(3x+4)^2} \\ &= \frac{4}{(3x+4)^2} \end{align*}
  1. Differentiate ex2x\frac{e^{x^2}}{x}
u=ex2u=ex22xv=xv=1dydx=2x2ex2ex2x2=ex2(2x21)x2\begin{align*} u &= e^{x^2} \\ u' &= e^{x^2} \cdot 2x \\ \\ v &= x \\ v' &= 1 \\ \\ \frac{dy}{dx} &= \frac{2x^2e^{x^2} - e^{x^2}}{x^2} \\ &= \frac{e^{x^2}(2x^2 - 1)}{x^2} \end{align*}

Conclusion

The content in this series is mostly based on the IGCSE Further Math Textbook, and the subsequent sections are accessible through this section at the top of the article.

Hopefully it was a decent starting point that eased you into the basics of calculus.

Thanks for reading.