# Differentiation by First Principles

## Finding Gradients of a Straight Line using First Principles

Suppose we are given a function $f(x)$ and asked to find its derivative at the point where $x=a$. This is actually the gradient of the tangent to the curve at $x=a$, which we write as $f^{\prime}(a)$.
There are two methods for finding $f^{\prime}(a)$ using first principles:

### Definition of the Gradient Function

$$f^{\prime}(a)=\lim_{h \rightarrow 0}\dfrac{f(a+h)-f(a)}{h}$$

### Example 1

Use the first principles formula $\displaystyle f'(a)=\lim_{h \rightarrow 0}\dfrac{f(a+h)-f(a)}{h}$ to find the instantaneous rate of change in $f(x)=x^2-2x$ at the point where $x=3$.

\begin{align} \displaystyle f(3) &= 3^2 -2 \times 3 = 3 \\ f^{\prime}(3) &= \lim_{h \rightarrow 0} \dfrac{f(3+h)-f(3)}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{(3+h)^2-2(3+h)-3}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{9+6h+h^2-6-2h-3}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{h^2+4h}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{(h+4)h}{h} \\ &= \lim_{h \rightarrow 0} (h+4) \\ &= 4 \end{align}

### Two Fixed Points of First Principles

The second method to consider two points on the graph of $y=f(x)$, a fixed point $\color{teal}A(a,f(a))$ and a variable point $\color{teal}B(x,f(x))$.

The gradient of chord $\color{teal}A\color{teal}B=\dfrac{f(x)-f(a)}{x-a}$.
In the limit as $\color{teal}B$ approaches $\color{teal}A$, $x \rightarrow a$ and the gradient of chord $\color{teal}A\color{teal}B$ approaches gradient of the tangent at $\color{teal}A$.
$$f^{\prime}(a)=\lim_{x \rightarrow a}\dfrac{f(x)-f(a)}{x-a}$$
This is an alternative definition of the gradient of the tangent at $x=a$.
Note that the gradient of the tangent at $x=a$ is defined as the gradient of the curve at the point where $x=a$, and is the instantaneous rate of change in $y$ with respect to $x$ at that point.

### Example 2

Use the first principles formula $\displaystyle f'(a)=\lim_{x \rightarrow a}\dfrac{f(x)-f(a)}{x-a}$ to find the instantaneous rate of change in $f(x)=2x^2+9$ at the point where $x=2$.

\begin{align} \displaystyle f(2) &= 2 \times 2^2 + 9 = 17 \\ f^{\prime}(2) &= \lim_{x \rightarrow 2} \dfrac{f(x)-f(2)}{x-2} \\ &= \lim_{x \rightarrow 2} \dfrac{2x^2+9 -17}{x-2} \\ &= \lim_{x \rightarrow 2} \dfrac{2x^2 – 8}{x-2} \\ &= \lim_{x \rightarrow 2} \dfrac{2(x^2 – 4)}{x-2} \\ &= \lim_{x \rightarrow 2} \dfrac{2(x – 2)(x+2)}{x-2} \\ &= \lim_{x \rightarrow 2} 2(x+2) \\ &= 2(2+2) \\ &= 8 \end{align}

## Differentiation by First Principles

Consider a function $y=f(x)$ where $\color{teal}A$ is the point $(x,f(x))$ and $\color{teal}B$ is the point $(x+h,f(x+h))$.

The chord $\color{teal}A\color{teal}B$ has gradient $\dfrac{f(x+h)-f(x)}{(x+h)-x} = \dfrac{f(x+h)-f(x)}{h}$.
The gradient of the tangent at the variable point $(x,f(x))$ is the limiting value of $\dfrac{f(x+h)-f(x)}{h}$ as $h$ approaches $0$.
This formula gives the gradient of the tangent to the curve $y=f(x)$ at the point $(x,f(x))$, for any value of the variable $x$ for which this limit exists. Since there is at most one value of the gradient for each value of $x$, the formula is actually a function.

The derivative of $y=f(x)$ is defined as;
$$f^{\prime}(x)=\lim_{h \rightarrow 0}\dfrac{f(x+h)-f(x)}{h}$$
For we evaluate this limit to find a derivative, we say we perform $\textit{Differentiation from First Principles}$.

## The Derivative when $x=a$

The gradient of the tangent to $y=f(x)$ at the point where $x=a$ is denoted $f^{\prime}(a)$,
$$f^{\prime}(a)=\lim_{h \rightarrow 0}\dfrac{f(a+h)-f(a)}{h}$$

### Fundamental Theory of First Principles

Use the definition of $f^{\prime}(x)$ to find the derivative of $f(x)=x^2$.

\begin{align} \displaystyle f^{\prime}(x) &= \lim_{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{(x+h)^2-x^2}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{x^2+2xh+h^2-x^2}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{2xh+h^2}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{(2x+h)h}{h} \\ &= \lim_{h \rightarrow 0} (2x+h) \\ &= 2x \end{align}

### Derivative by First Principles of Quadratics involving Coefficients

Use the definition of $f^{\prime}(x)$ to find the derivative of $f(x)=3x^2$.

\begin{align} \displaystyle f^{\prime}(x) &= \lim_{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{3(x+h)^2-3x^2}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{3x^2+6xh+3h^2-3x^2}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{6xh+3h^2}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{(6x+3h)h}{h} \\ &= \lim_{h \rightarrow 0} (6x+3h) \\ &= 6x \end{align}

### Derivative by First Principles of Quadratics involving Constants

Use the definition of $f^{\prime}(x)$ to find the derivative of $f(x)=x^2+1$.

\begin{align} \displaystyle f^{\prime}(x) &= \lim_{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{((x+h)^2+1)-(x^2+1)}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{x^2+2xh+h^2+1-x^2-1}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{2xh+h^2}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{(2x+h)h}{h} \\ &= \lim_{h \rightarrow 0} (2x+h) \\ &= 2x \end{align}

### Derivative by First Principles of Quadratics involving Linear Terms

Use the definition of $f^{\prime}(x)$ to find the derivative of $f(x)=x^2-x$.

\begin{align} \displaystyle f^{\prime}(x) &= \lim_{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{((x+h)^2-(x+h))-(x^2-x)}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{x^2+2xh+h^2-x-h-x^2+x}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{2xh+h^2}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{(2x+h)h}{h} \\ &= \lim_{h \rightarrow 0} (2x+h) \\ &= 2x \end{align}

### Derivative by First Principles of Cubic Expressions

Use the definition of $f^{\prime}(x)$ to find the derivative of $f(x)=x^3$.

\begin{align} \displaystyle f^{\prime}(x) &= \lim_{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{(x+h)^3-x^3}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{x^3+3x^2h+3xh^2+h^3-x^3}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{3x^2h+3xh^2+h^3}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{(3x^2+3xh+h^2)h}{h} \\ &= \lim_{h \rightarrow 0} (3x^2+3xh+h^2) \\ &= 3x^2 \end{align}

### Derivative by First Principles of Square Roots

Use the definition of $f^{\prime}(x)$ to find the derivative of $f(x)=\sqrt{x}$.

\begin{align} \displaystyle \require{color} f^{\prime}(x) &= \lim_{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{\sqrt{x+h}-\sqrt{x}}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{\sqrt{x+h}-\sqrt{x}}{h} \times \dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} &\require{AMSsymbols} \color{red}{\text{rationalise the numerator}} \\ &= \lim_{h \rightarrow 0} \dfrac{\sqrt{x+h}^2-\sqrt{x}^2}{h(\sqrt{x+h}+\sqrt{x})} \\ &= \lim_{h \rightarrow 0} \dfrac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})} \\ &= \lim_{h \rightarrow 0} \dfrac{h}{h(\sqrt{x+h}+\sqrt{x})} \\ &= \lim_{h \rightarrow 0} \dfrac{1}{\sqrt{x+h}+\sqrt{x}} \\ &= \dfrac{1}{\sqrt{x}+\sqrt{x}} \\ &= \dfrac{1}{2\sqrt{x}} \\ \end{align}

### Derivative by First Principles of Square Roots with Coefficients and Constants

Differentiate $f(x) = \sqrt{5x+6}$ from first principles.

\begin{aligned} \displaystyle f'(x) &= \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \\ &= \lim_{h\to 0} \frac{\sqrt{5(x+h)+6} – \sqrt{5x+6}}{h} \\ &= \lim_{h\to 0} \frac{\sqrt{5x+5h+6} – \sqrt{5x+6}}{h} \times \frac{\sqrt{5x+5h+6} + \sqrt{5x+6}}{\sqrt{5x+5h+6} + \sqrt{5x+6}} \\ &= \lim_{h\to 0} \frac{(5x+5h+6)-(5x+6)}{h(\sqrt{5x+5h+6} + \sqrt{5x+6})} \\ &= \lim_{h\to 0} \frac{5h}{h(\sqrt{5x+5h+6} + \sqrt{5x+6})} \\ &= \lim_{h\to 0} \frac{5}{\sqrt{5x+5h+6} + \sqrt{5x+6}} \\ &= \frac{5}{\sqrt{5x+6} + \sqrt{5x+6}} \\ &= \frac{5}{2\sqrt{5x+6}} \end{aligned}

### Derivative by First Principles of Rational Expressions

Use the definition of $f^{\prime}(x)$ to find the derivative of $f(x)=\dfrac{1}{x}$.

\begin{align} \displaystyle \require{color} f^{\prime}(x) &= \lim_{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{\dfrac{1}{x+h}-\dfrac{1}{x}}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{\dfrac{x-(x+h)}{(x+h)x}}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{\dfrac{-h}{(x+h)x}}{h} \\ &= \lim_{h \rightarrow 0} \dfrac{-h}{(x+h)hx} \\ &= \lim_{h \rightarrow 0} \dfrac{-1}{(x+h)x} \\ &= -\dfrac{1}{x \times x} \\ &= -\dfrac{1}{x^2} \end{align}

### Derivative by First Principles of Cube Roots

Differentiate $\displaystyle f(x) = \sqrt[3]{x}$ from first principles.

\begin{aligned} \displaystyle f'(x) &= \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \\ &= \lim_{h\to 0} \frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h} \\ &= \lim_{h\to 0} \frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h} \times \frac{\sqrt[3]{(x+h)^2} + \sqrt[3]{x+h} \times \sqrt[3]{x} + \sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x+h} \times \sqrt[3]{x} + \sqrt[3]{x^2}} \\ &= \lim_{h\to 0} \frac{\sqrt[3]{x+h}^3-\sqrt[3]{x}^3}{h\big(\sqrt[3]{(x+h)^2} + \sqrt[3]{x+h} \times \sqrt[3]{x} + \sqrt[3]{x^2}\big)} \\ &= \lim_{h\to 0} \frac{x+h-x}{h\big(\sqrt[3]{(x+h)^2} + \sqrt[3]{x+h} \times \sqrt[3]{x} + \sqrt[3]{x^2}\big)} \\ &= \lim_{h\to 0} \frac{h}{h\big(\sqrt[3]{(x+h)^2} + \sqrt[3]{x+h} \times \sqrt[3]{x} + \sqrt[3]{x^2}\big)} \\ &= \lim_{h\to 0} \frac{1}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x+h} \times \sqrt[3]{x} + \sqrt[3]{x^2}} \\ &= \frac{1}{\sqrt[3]{x^2} + \sqrt[3]{x} \times \sqrt[3]{x} + \sqrt[3]{x^2}} \\ &= \frac{1}{3\sqrt[3]{x^2}} \end{aligned}

## Frequently Asked Questions

### What is the formula of differentiation by first principles?

$\displaystyle f^{\prime}(x)=\lim_{h \rightarrow 0}\dfrac{f(x+h)-f(x)}{h}$

### What is the definition of differentiation by first principles?

Differentiation by first principles refers to find a general expression for the slope or gradient of a curve using algebraic techniques. The derivative using is a measure of the instantaneous rates of change, which is the gradient of a specific point of the curve.

### How do you differentiate by first principles?

Let’s locate two fixed points in a curve, say A and B. The rate of the change of vertical values and the horizontal value refers to the slope of a straight line connecting two points. The limit value of the rate when the horizontal distance of these two points are approaching to zero is referred to as instantaneous rates of change and the technique being used is called “differentiation by first principles”.