First Principle of Derivatives – Definition, Formula, and Applications

In JEE Mathematics, a strong understanding of calculus is critical to problem-solving limits, continuity, differentiation and integration. The First Principle of Derivatives is the basis on which all differentiation rules are developed.

Most students simply use the power rule, product rule or chain rule of differentiation but when it comes to exams like JEE Main and Advanced conceptual questions, understanding first principle differentiation is critical to a deep understanding of the first principle approach.

In this guide you will be able to find the definition, formula, derivations, solved examples and common pitfalls to avoid in order to be prepared for your exam.

1.0What is the First Principle of Derivatives?

First Principle of Derivatives Definition

The First Principle of Derivatives talks about how to use limits to define a derivative. It says that the derivative of a function at a point is the rate at which the function changes at that moment with regard to the independent variable.

Mathematically:

• If f(x) is a real function, the derivative of f at point x is given by the limit of the difference quotient as the change in input approaches zero.

In simple words:
The first principle is the basic method of calculating the derivative using limits rather than short cuts. The derivative informs us how quickly a function changes.

2.0The Formula of the First Principle of Derivatives

General Formula

The First Principle of Derivatives formula is:

$f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$

Where:

• f′(x) = derivative of f(x) at x
• h = small increment in x

This formula is also called the definition of derivative or Newton’s quotient.

Step-by-Step Derivation

1. Suppose we want to find the slope of a curve y=f(x) at a point x.
2. Consider a nearby point x+h.
3. The slope of the secant line between these two points is: $\frac{f(x+h)-f(x)}{h}$
   
4. As h approaches zero, the secant line becomes the tangent line.
5. Therefore, the slope of the tangent line is:
   $f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$

This is the fundamental derivative definition.

3.0Proof of First Principles of Derivatives

Let’s prove the formula step by step:

1. Consider a curve y=f(x).
2. Suppose we want to find the slope of the curve at point P(x,f(x))
3. Take another point Q(x+h, f(x+h)).
4. The slope of the secant line PQ is: $m_{PQ}=\frac{f(x+h)-f(x)}{h}$
   
5. As Q approaches P, i.e., h→0 the secant becomes the tangent.
6. The slope of the tangent is therefore:$m_{\text{tangent}}={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$
7. This slope is defined as the derivative of f(x) at x.

Hence,

$f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$

This completes the proof of the First Principle of Derivatives.

4.0How to Find Derivatives using the First Principle?

Finding derivatives using the first principle requires substitution, expansion, and simplification.

Step-by-step Process

1. Write down the given function f(x).
2. Substitute it into the formula:
   $f^{\prime }(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$
3. Simplify the numerator (expand using binomial or trigonometric identities).
4. Cancel out common terms and factorize wherever possible.
5. Apply limit properties to evaluate as h→0.

5.0Application of the First Principle of Derivatives

Derivative of Standard Functions

Let’s derive derivatives of some standard functions using the First Principle of Derivatives formula.

(i) Derivative of $f(x)=x^n$

$f^{\prime }(x)={\lim }_{h\to 0}\frac{(x+h{)}^n-x^n}{h}$

Using Binomial Expansion:

$(x+h{)}^n=x^n+n{x}^{n-1}h+\frac{n(n-1)}{2}{x}^{n-2}{h}^2+\dots$

So,

$f^{\prime }(x)={\lim }_{h\to 0}\frac{n{x}^{n-1}h+\frac{n(n-1)}{2}{x}^{n-2}{h}^2+\dots }{h}$

$f^{\prime }(x)={\lim }_{h\to 0}\left[n{x}^{n-1}+\frac{n(n-1)}{2}{x}^{n-2}h+\dots \right]$

$f^{\prime }(x)=n{x}^{n-1}$

Verified formula.

(ii) Derivative of f(x)=sin⁡x

$f^{\prime }(x)={\lim }_{h\to 0}\frac{\sin (x+h)-\sin (x)}{h}$

Using identity:

$\sin (x+h)=\sin x\cos h+\cos x\sin h$

So,

$f^{\prime }(x)={\lim }_{h\to 0}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}$

Using limits:

$={\lim }_{h\to 0}\frac{\sin x(\cos h-1)+\cos x\sin h}{h}$

${\lim }_{h\to 0}\frac{\sin h}{h}=1$

So,

f′(x)=cos⁡x

(iii) Derivative of f(x)=cos⁡x

Similar process gives:

f′(x)=−sin⁡x

(iv) Derivative of $f(x)=e^x$

$f^{\prime }(x)={\lim }_{h\to 0}\frac{e^{x+h}-e^x}{h}$

$=e^x{\lim }_{h\to 0}\frac{e^h-1}{h}$

$=e^x$

6.0Solved Example 

Example 1: Using the First Principle of Derivatives, find $\frac{d}{dx}(\ln x).$

Solution:

$f^{\prime }(x)={\lim }_{h\to 0}\frac{\ln (x+h)-\ln x}{h}$

$={\lim }_{h\to 0}\frac{\ln \left(\frac{x+h}{x}\right)}{h}$

$={\lim }_{h\to 0}\frac{\ln \left(1+\frac{h}{x}\right)}{h}$

Put $k=\frac{h}{x}⟹h=kx:$

$f^{\prime }(x)={\lim }_{k\to 0}\frac{\ln (1+k)}{kx}$

$=\frac{1}{x}{\lim }_{k\to 0}\frac{\ln (1+k)}{k}$

Since ${\lim }_{k\to 0}\frac{\ln (1+k)}{k}=1,$

$f^{\prime }(x)=\frac{1}{x}$

Example 2: Derivative of $f(x)=x^2$

$f^{\prime }(x)={\lim }_{h\to 0}\frac{(x+h{)}^2-x^2}{h}$

$={\lim }_{h\to 0}\frac{x^2+2xh+h^2-x^2}{h}$

$={\lim }_{h\to 0}\frac{2xh+h^2}{h}$

$={\lim }_{h\to 0}(2x+h)=2x$

$\frac{d}{dx}(x^2)=2x.$

Example 3: Derivative of f(x)=sin⁡x

$f^{\prime }(x)={\lim }_{h\to 0}\frac{\sin (x+h)-\sin x}{h}$

$\text{Using identity: }\sin (A+B)=\sin A\cos B+\cos A\sin B$

$={\lim }_{h\to 0}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}$

$={\lim }_{h\to 0}\frac{\sin x(\cos h-1)+\cos x\sin h}{h}$

Using limits:

${\lim }_{h\to 0}\frac{\sin h}{h}=1$

${\lim }_{h\to 0}\frac{\cos h-1}{h}=0$

$f^{\prime }(x)=\cos x$

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

7.0Practice Questions 

1. Using the First Principle of Derivatives, prove that $\frac{d}{dx}(x^3)=3x^2$
2. Find the derivative of f(x) = tan⁡x using the first principle. $f(x)=1/xis-1/x^2$
3. Using first principle, show that the derivative of
4. Evaluate ${\lim }_{h\to 0}\frac{(a+h{)}^n-a^n}{h}$ and interpret its meaning.
5. Using first principle, prove that $\frac{d}{dx}(\sin 2x)=2\cos 2x$

8.0Why Study the First Principle of Derivatives for JEE?

• Conceptual Clarity: Basic comprehension is frequently tested in JEE.
• The first principle is used to derive all derivative formulas, such as $d(xn)/dx=nxn-1$, as proof of standard results.
• Advanced Topics: The first principle is expanded upon by ideas such as error analysis, Taylor series, and approximations.
• Advanced JEE: Instead of using the normal rule, you can be asked to compute a derivative using the first principle