Proof of Difference rule of Differentiation

The difference rule of derivatives is actually derived in differential calculus from first principle. For example, $f{(x)}$ and $g{(x)}$ are two differentiable functions and the difference of them is written as $f{(x)}-g{(x)}$. The derivative of difference of two functions with respect to $x$ is written in the following mathematical form.

$\dfrac{d}{dx}{\, \Big(f{(x)}-g{(x)}\Big)}$

Define the derivative in limiting operation

Take $m{(x)} = f{(x)}-g{(x)}$ and then $m{(x+\Delta x)} = f{(x+\Delta x)}-g{(x+\Delta x)}$

According to definition of the derivative, write the derivative of the function $m{(x)}$ with respect to $x$ in limiting operation.

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

Replace the actual functions of $m{(x)}$ and $m{(x+\Delta x)}$.

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

Simplify the function

Now, take $\Delta x = h$ and start simplifying this function for deriving the derivative of difference of two functions by first principle.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(f{(x+h)}-g{(x+h)}\Big)-\Big(f{(x)}-g{(x)}\Big)}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f{(x+h)}-g{(x+h)}-f{(x)}+g{(x)}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f{(x+h)}-f{(x)}-\Big(g{(x+h)}-g{(x)}\Big)}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{f{(x+h)}-f{(x)}}{h}-\dfrac{g{(x+h)}-g{(x)}}{h}\Bigg]}$

Evaluate the Limits of the functions

As per difference rule of limits, the limit of difference of two functions can be written as difference of their limits.

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f{(x+h)}-f{(x)}}{h}}$ $-$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{g{(x+h)}-g{(x)}}{h}}$

According to first principle of differentiation, each term in the right right-hand side of the equation represents the derivative of the respective function.

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, \Big(f{(x)}-g{(x)}\Big)}$ $\,=\,$ $\dfrac{d}{dx}{\, f{(x)}}$ $-$ $\dfrac{d}{dx}{\, g{(x)}}$

In this way, the difference rule of derivatives can be derived in differential calculus mathematically from first principle.

The derivative difference rule is also written in two forms alternatively by taking $u = f{(x)}$ and $v = g{(x)}$.

Leibniz’s notation

$(1) \,\,\,$ $\dfrac{d}{dx}{\, (u-v)}$ $\,=\,$ $\dfrac{du}{dx}$ $-$ $\dfrac{dv}{dx}$

Differentials notation

$(2) \,\,\,$ ${d}{\, (u-v)}$ $\,=\,$ $du-dv$

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved