Math Doubts

Proof of Sum Rule of Differentiation

The sum rule of differentiation can be derived in differential calculus from first principle. $f{(x)}$ and $g{(x)}$ are two differential functions and the sum of them is written as $f{(x)}+g{(x)}$. The derivative of sum of two functions with respect to $x$ is expressed in mathematical form as follows.

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

Define the derivative in limiting operation

Take, $s{(x)} = f{(x)}+g{(x)}$ and then $s{(x+\Delta x)} = f{(x+\Delta x)}+g{(x+\Delta x)}$

Now, express the derivative of the function $s{(x)}$ with respect to $x$ in limiting operation as per definition of the derivative.

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

Now, replace the values of functions $s{(x)}$ and $s{(x+\Delta x)}$

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

Simplify the function

Take $\Delta x = h$ and simplify the function for deriving the derivative of sum of the functions.

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

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

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

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

Evaluate the Limits of the functions

The limit of sum of functions is equal to sum of their limits as per sum rule of limits.

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

According to the first principle of derivative, the limit of each function is the derivative of that function.

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

Therefore, it is proved that the derivative of sum of two functions is equal to sum of their derivatives. The derivative property is called the sum rule of differentiation. The derivative sum rule can also be used to sum of more than two terms.

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

Thus, the sum rule of differentiation is derived mathematically in differential calculus from first principle.

Take $u = f{(x)}$, $v = g{(x)}$ and so on. The sum rule of derivatives can be written in two forms.

Leibniz’s notation

$(1) \,\,\,$ $\dfrac{d}{dx}{\, (u+v+w+\ldots)}$ $\,=\,$ $\dfrac{du}{dx}$ $+$ $\dfrac{dv}{dx}$ $+$ $\dfrac{dw}{dx}$ $+$ $\ldots$

Differentials notation

$(2) \,\,\,$ ${d}{\, (u+v+w+\ldots)}$ $\,=\,$ $du$ $+$ $dv$ $+$ $dw$ $+$ $\ldots$



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