First Principle of Derivatives

First Principle of Differentiation involves finding the derivative of a function using the fundamental definition of the derivative. This method requires calculating the limit of the difference quotient as the interval between two points on the function approaches zero.

In this article, we will learn about the first principle of derivative, its definition, its proof, how to find derivatives using the first principle, one-sided derivative and solved examples for better understanding.

First Principles of Derivatives Definition

The first principle of derivatives involves using algebra to determine a general expression for the slope of a curve. It is also referred to as the delta method. The derivative serves as a measure of the instantaneous rate of change, denoted by f'(x), which is equal to:

[Tex]f'(x) = \frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}[/Tex]

Where,

• f′(x) represents the derivative of the function f(x) with respect to x.
• h represents the change in the x-values between the two points.
• f(x+h)−f(x) represents the change in the y-values between the two points.
• The limit as h approaches zero ensures that the secant line becomes the tangent line, providing the instantaneous rate of change of the function at the point x.

Proof of First Principles of Derivatives

Given a function f(x), we want to find its derivative f'(x). Using the definition of the derivative, we have:

[Tex]f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}[/Tex]

Consider two points on the curve f(x) at x = a and x = a + h. The slope of the secant line passing through these points is given by:

[Tex] \text{Slope} = \frac{f(a + h) – f(a)}{(a + h) – a} = \frac{f(a + h) – f(a)}{h} [/Tex]

As h approaches 0, the secant line becomes a tangent line, and its slope represents the derivative of the function at x = a .

Taking the limit of the difference quotient as h approaches 0 gives us the derivative at x = a:

[Tex] f'(a) = \lim_{h \to 0} \frac{f(a + h) – f(a)}{h} [/Tex]

This limit represents the instantaneous rate of change of the function f(x) at x = a .

How to Find Derivative using the First Principle?

To find the derivative of a function using the first principles, follow these steps:

Step 1: Start with the function (f(x)) for which you want to find the derivative.

Step 2: Use the definition of the derivative: [Tex]f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}[/Tex]

Step 3: Substitute f(x) into the formula.

Step 4: Calculate (f(x+h) – f(x)), which represents the change in ( y )-values between two points on the function.

Step 5: Divide the result by ( h ), the change in ( x )-values between the two points.

Step 6: Take the limit as ( h ) approaches zero.

Step 7: The resulting value is the derivative of f(x) with respect to ( x ), denoted as f'(x).

One-Sided Derivative

One-sided derivatives are derivatives calculated from one direction only, either from the left or the right of a specific point. They are useful when a function behaves differently on one side of the point compared to the other.

• Left Hand Derivative
• Right Hand Derivative

Left-sided derivative [f’_–(x)]

This represents the derivative from the left side of a point (x). It is calculated by taking the limit as (h) approaches zero from the left side of (x).

[Tex]f’_-(x) = \lim_{h \to 0^-} \frac{f(x) – f(x – h)}{h} [/Tex]

Right-sided derivative [f’₊(x)]

This represents the derivative from the right side of a point (x). It is calculated by taking the limit as (h) approaches zero from the right side of (x).

[Tex]f’_+(x) = \lim_{h \to 0^+} \frac{f(x + h) – f(x)}{h}[/Tex]

Example: Consider ( f(x) = x² ) and find the one-sided derivatives at ( x = 1 ).

Solution:

Left-sided derivative ( f’_–(1)):

f’_–(1) = lim_h→0- [Tex]\frac{f(1) – f(1 – h)}{h} [/Tex]

f’_–(1) = lim_h→0- [Tex]\frac{(1)^2 – (1 – h)^2}{h}[/Tex]

f’_–(1) = lim_h→0- [Tex]\frac{1 – (1 – 2h + h^2)}{h}[/Tex]

f’_–(1) = lim_h→0- [Tex]\frac{1 – 1 + 2h – h^2}{h}[/Tex]

f’_–(1) = lim_h→0- [Tex]\frac{2h – h^2}{h}[/Tex]

f’_–(1) = lim_h→0- (2 + h) = 2

Right-sided derivative ( f’₊(1)):

f’₊(1) = lim_h→0+ [Tex]\frac{f(1 + h) – f(1)}{h}[/Tex]

f’₊(1) = lim_h→0+ [Tex]\frac{(1 + h)^2 – (1)^2}{h}[/Tex]

f’₊(1) = lim_h→0+ [Tex]\frac{(1 + 2h + h^2) – 1}{h}[/Tex]

f’₊(1) = lim_h→0+ [Tex]\frac{2h + h^2}{h}[/Tex]

f’₊(1) = lim_h→0+ (2 + h) = 2

So, both the left-sided derivative ( f’_–(1)) and the right-sided derivative ( f’₊(1)) of ( f(x) = x² ) at ( x = 1 ) are equal to 2.

Differentiation of  Functions Using First Principles of Derivatives

To differentiate trigonometric functions using the first principles of derivatives, use the definition of the derivative: [Tex]f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}[/Tex]

• Derivation of sin x: = cos x
• Derivative of cos x: = -sin x
• Derivative of tan x: = sec²x
• Derivative of cot x: = −cosec²x
• Derivative of sec x: = sec x.tan x
• Derivative of cosec x: = -cosec x.cot x

Derivative of Sinx by First Principle

Given: f(x) = sin(x)

Using the definition of the derivative:

[Tex]\frac{d}{dx}[/Tex] sin(x) = lim_{h → 0} [Tex]\frac{\sin(x + h) – \sin(x)}{h} [/Tex]

Applying the angle addition formula for sine:

lim_{h → 0} [Tex]\frac{\sin(x)\cos(h) + \cos(x)\sin(h) – \sin(x)}{h}[/Tex]

lim_{h → 0} [Tex]\frac{\sin(x)(\cos(h) – 1) + \cos(x)\sin(h)}{h}[/Tex]

Using the limits:

sin(x) lim_{h → 0} [Tex]\frac{\cos(h) – 1}{h}[/Tex] + cos(x) lim_{h → 0} [Tex]\frac{\sin(h)}{h} [/Tex]

As lim_{h → 0} [Tex]\frac{\sin(h)}{h}[/Tex] = 1 and lim_{h → 0} [Tex]\frac{\cos(h) – 1}{h}[/Tex] = 0 :

sin(x) · 0 + cos(x) · 1 = cos(x)

So, the derivative of sin(x) with respect to ( x ) using the first principles of derivatives is cos(x).

Derivative of Cosx by First Principle

[Tex]\frac{d}{dx}(\cos(x)[/Tex] = lim_{h → 0} [Tex]\frac{\cos(x + h) – \cos(x)}{h} [/Tex]

Using the angle addition formula for cosine:

= lim_{h → 0} [Tex]\frac{\cos(x)\cos(h) – \sin(x)\sin(h) – \cos(x)}{h}[/Tex]

= lim_{h → 0} [Tex]\frac{\cos(x)(\cos(h) – 1) – \sin(x)\sin(h)}{h}[/Tex]

= cos(x) lim_{h → 0} [Tex]\frac{\cos(h) – 1}{h}[/Tex] – sin(x) lim_{h → 0} [Tex]\frac{\sin(h)}{h}[/Tex]

As lim_{h → 0} [Tex]\frac{\sin(h)}{h}[/Tex] = 1 and lim_{h → 0} [Tex]\frac{\cos(h) – 1}{h}[/Tex] = 0 \):

= cos(x) ⋅ 0 – sin(x) ⋅ 1 = -sin(x)

So, the derivative of cos(x) with respect to ( x ) using the first principles of derivatives is -sin(x).

Derivative of Tanx by First Principle

[Tex] \frac{d}{dx}(\tan(x)) = \lim_{h \to 0} \frac{\tan(x + h) – \tan(x)}{h}[/Tex]

Using the tangent addition formula:

= [Tex]\lim_{h \to 0} \frac{\frac{\sin(x + h)}{\cos(x + h)} – \frac{\sin(x)}{\cos(x)}}{h} [/Tex]

= [Tex]\lim_{h \to 0} \frac{\sin(x + h)\cos(x) – \sin(x)\cos(x + h)}{h\cos(x)\cos(x + h)} [/Tex]

= [Tex]\lim_{h \to 0} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) – \sin(x)\cos(x)}{h\cos(x)\cos(h)} [/Tex]

= [Tex]\lim_{h \to 0} \frac{\sin(x)(\cos(h) – \cos(x)) + \cos(x)\sin(h)}{h\cos(x)\cos(h)}[/Tex]

= [Tex]\frac{\sin(x)(1 – \cos(x))}{\cos^2(x)} + \frac{\cos(x)}{\cos^2(x)} \lim_{h \to 0} \frac{\sin(h)}{h}[/Tex]

= [Tex]\frac{\sin(x)(1 – \cos(x))}{\cos^2(x)} + \frac{\cos(x)}{\cos^2(x)} \cdot 1[/Tex]

= [Tex]\frac{\sin(x) – \sin(x)\cos(x) + \cos(x)}{\cos^2(x)}[/Tex]

= [Tex]\frac{\sin(x) + \cos(x) – \sin(x)\cos(x)}{\cos^2(x)}[/Tex]

= [Tex]\frac{\sin(x) + \cos(x)}{\cos^2(x)} – \frac{\sin(x)\cos(x)}{\cos^2(x)}[/Tex]

= [Tex]\frac{\sin(x)}{\cos^2(x)} + \frac{\cos(x)}{\cos^2(x)} – \tan(x)\sec^2(x)[/Tex]

= [Tex]\frac{\sin(x) + \cos(x)}{\cos^2(x)} – \tan(x)\sec^2(x)[/Tex]

= [Tex]\frac{\sin(x) + \cos(x)}{\cos^2(x)} – \frac{\sin(x)}{\cos^2(x)}[/Tex]

= [Tex]\frac{\cos(x)}{\cos^2(x)}[/Tex]

= [Tex]\frac{1}{\cos(x)}[/Tex]

= sec(x)

So, the derivative of tan(x) with respect to ( x ) using the first principles of derivatives is sec(x).

Solved Examples

Example 1: Find the derivative of the function f(x) = 3x² + 2x – 1 using the first principles of differentiation.

Solution:

Using the definition of the derivative:

f'(x) = lim_{h → 0} \frac{f(x + h) – f(x)}{h}

Substituting \( f(x) = 3x^2 + 2x – 1 \) into the formula:

f'(x) = lim_{h → 0} [Tex]\frac{3(x + h)^2 + 2(x + h) – 1 – (3x^2 + 2x – 1)}{h}[/Tex]

f'(x) = lim_{h → 0} [Tex]\frac{3(x^2 + 2xh + h^2) + 2x + 2h – 1 – 3x^2 – 2x + 1}{h}[/Tex]

f'(x) = lim_{h → 0} [Tex]\frac{3x^2 + 6xh + 3h^2 + 2x + 2h – 1 – 3x^2 – 2x + 1}{h} [/Tex]

f'(x) = lim_{h → 0} [Tex]\frac{6xh + 3h^2 + 2h}{h} [/Tex]

f'(x) = lim_{h → 0} (6x + 3h + 2)

f'(x) = 6x + 2

So, the derivative of f(x) = 3x² + 2x – 1 with respect to ( x ) using the first principles of differentiation is f'(x) = 6x + 2.

Example 2: Calculate the derivative of the function ( g(x) = √x) using the first principles of differentiation.

Solution:

Using the definition of the derivative:

g'(x) = lim_{h → 0} [Tex]\frac{g(x + h) – g(x)}{h}[/Tex]

Substituting (g(x) = √x) into the formula:

g'(x) = lim_{h → 0} [Tex]\frac{\sqrt{x + h} – \sqrt{x}}{h}[/Tex]

To simplify this expression, use the conjugate:

g'(x) = lim_{h → 0} [Tex]\frac{\sqrt{x + h} – \sqrt{x}}{h} \times \frac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}} [/Tex]

g'(x) = lim_{h → 0} [Tex]\frac{(x + h) – x}{h(\sqrt{x + h} + \sqrt{x})}[/Tex]

g'(x) = lim_{h → 0} \frac{h}{h(\sqrt{x + h} + \sqrt{x})} \]

g'(x) = lim_{h → 0} [Tex]\frac{1}{\sqrt{x + h} + \sqrt{x}}[/Tex]

Now, as ( h ) approaches 0:

g'(x) = [Tex]\frac{1}{2\sqrt{x}} [/Tex]

So, the derivative of ( g(x) = √x) with respect to ( x ) using the first principles of differentiation is ( g'(x) = [Tex]\frac{1}{2\sqrt{x}} [/Tex]).

Example 3: Determine the derivative of the function h(x) = e^x using the first principles of differentiation.

Solution:

Using the definition of the derivative:

h'(x) = e^x · lim_{h → 0} [Tex]\frac{h(x + h) – h(x)}{h}[/Tex]

Substituting \( h(x) = e^x \) into the formula:

h'(x) = e^x · lim_{h → 0} [Tex]\frac{e^{x + h} – e^x}{h} [/Tex]

h'(x) = e^x · lim_{h → 0} [Tex]\frac{e^x \cdot e^h – e^x}{h}[/Tex]

h'(x) = e^x · lim_{h → 0} \[Tex]frac{e^x (e^h – 1)}{h} [/Tex]

Now, as ( h ) approaches 0:

h'(x) = e^x · lim_{h → 0} [Tex]\frac{e^h – 1}{h}[/Tex]

Using the definition of the derivative of ( e^x ) at ( x = 0 ), which is equal to 1:

h'(x) = e^x · 1 = e^x

So, the derivative of ( h'(x) = e^x) with respect to ( x ) using the first principles of differentiation is (h'(x) = e^x).

Practice Questions: First Principle of Derivatives

Q1. Find the derivative of the function ( f(x) = x³ + 2x² – 3x + 1 ) using the first principles of differentiation.

Q2. Calculate the derivative of the function ( g(x) = 1/x) using the first principles of differentiation.

Q3. Determine the derivative of the function ( h(x) = \ln(x) ) using the first principles of differentiation.

Q4. Find the derivative of the function ( p(x) = 1/√x) using the first principles of differentiation.

Q5. Calculate the derivative of the function ( q(x) = e^2x) using the first principles of differentiation.

First Principle of Derivatives: FAQs

What is first principle derivatives?

First principle derivatives refer to finding the derivative of a function using the basic definition of the derivative. It involves calculating the limit of the difference quotient as the interval approaches zero.

How do you find the first principles?

To find the first principles of derivatives, apply the definition of the derivative, which involves taking the limit of the slope of a secant line between two points on the function as the interval between the points approaches zero.

What is the second principle of the derivatives?

There is no commonly recognized “second principle of derivatives” in calculus. The primary principle is the definition of the derivative, which involves the limit of the difference quotient.

What is the derivative of 0?

The derivative of a constant function, such as f(x)=0, is always zero. This is because the slope of a horizontal line is zero at every point.