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 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
$$ \large f^{\prime}(a)=\lim_{h \rightarrow 0}\dfrac{f(a+h)-f(a)}{h}$$
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 tangent gradient 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 $y$ concerning $x$ 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 a function.
The derivative of $y=f(x)$ is defined as;
$$f^{\prime}(x)=\lim_{h \rightarrow 0}\dfrac{f(x+h)-f(x)}{h}$$
We perform $\textit{Differentiation from First Principles}$ to evaluate this limit to find a derivative.
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 finding a general expression for the slope or gradient of a curve using algebraic techniques. The derivative measures the instantaneous rates of change, which is a specific curve point gradient.
How do you differentiate by first principles?
Let’s locate two fixed points in a curve, A and B. The rate of the change of vertical and horizontal values 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 is approaching zero is referred to as instantaneous rates of change, and the technique being used is called “differentiation by first principles”.
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I enjoyed the tutorial, very nice and simple to understand, greate Job dear