Math Doubts

Proof of Derivative of a variable

Assume, $x$ is a variable. The derivative of a variable $x$ with respect to $x$ is written in mathematical form as follows in differential calculus.

$\dfrac{d}{dx}{\, (x)}$

Derivative of function in Limits form

Use definition of the derivative to express the differentiation of a function $f{(x)}$ with respect to $x$ in limits form. It is useful to prove the differentiation of variable $x$ by first principle.

$\dfrac{d}{dx}{\, f{(x)}}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to \, 0}{\normalsize \dfrac{f{(x+\Delta x)}-f{(x)}}{\Delta x}}$

Take $f{(x)} \,=\, x$, then $f{(x+\Delta x)} \,=\, x+\Delta x$. Now, replace them in the above formula.

$\implies$ $\dfrac{d}{dx}{\, (x)}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to \, 0}{\normalsize \dfrac{x+\Delta x-x}{\Delta x}}$

Take $\Delta x = h$ and write the equation in terms of $h$ in stead of $\Delta x$.

$\implies$ $\dfrac{d}{dx}{\, (x)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize \dfrac{x+h-x}{h}}$

$\implies$ $\dfrac{d}{dx}{\, (x)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize \dfrac{x-x+h}{h}}$

Evaluate the Derivative of variable

In numerator, there are three terms but two of them have opposite signs. So, they both get cancelled as per subtraction of the terms.

$\implies$ $\dfrac{d}{dx}{\, (x)}$ $\,=\,$ $\require{cancel} \displaystyle \large \lim_{h \,\to \, 0}{\normalsize \dfrac{\cancel{x}-\cancel{x}+h}{h}}$

$\implies$ $\dfrac{d}{dx}{\, (x)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize \Big(\dfrac{h}{h}\Big)}$

$\implies$ $\dfrac{d}{dx}{\, (x)}$ $\,=\,$ $\require{cancel} \displaystyle \large \lim_{h \,\to \, 0}{\normalsize \Big(\dfrac{\cancel{h}}{\cancel{h}}\Big)}$

The quotient of $h$ by $h$ is equal to one mathematically.

$\implies$ $\dfrac{d}{dx}{\, (x)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize \Big(1\Big)}$

There is no $h$ term in function. So, the limit of one as $h$ approaches zero is equal to one.

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, (x)}$ $\,=\,$ $1$

The derivative of a variable rule is derived from first principle in this way in differential calculus.

Math Doubts

A best free mathematics education website for students, teachers and researchers.

Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Maths Problems

Learn how to solve the maths problems in different methods with understandable steps.

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved