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.



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