# Proof of Product Rule of Derivatives

$f{(x)}$ and $g{(x)}$ are two differential functions in terms of $x$. The product of them is simply written as $f{(x)}.g{(x)}$ mathematically and the derivative of product of them with respect to $x$ is written in differential mathematics as follows.

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

Take $m{(x)} = {f{(x)}}.{g{(x)}}$

$\implies$ $\dfrac{d}{dx}{\, \Big({f{(x)}}.{g{(x)}}\Big)}$ $\,=\,$ $\dfrac{d}{dx}{\, m{(x)}}$

### Use Definition of the Derivative

The derivative of the product can be expressed in limiting operation form from first principle.

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

Actually, $m{(x)} = {f{(x)}}.{g{(x)}}$, then $m{(x+h)} = {f{(x+h)}}.{g{(x+h)}}$. Now, substitute them in the definition of the derivative in limiting operation form.

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

### A small acceptable Adjustment

Add and subtract $f{(x)}g{(x+h)}$ in the numerator of the function for factoring the mathematical expression in the numerator.

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

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

### Simplify the mathematical expression

The function $g{(x+h)}$ is a common factor in the first two terms and $f{(x)}$ is a common factor in the next two terms. Therefore, factorize the mathematical expression for expressing it in simple form.

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

Now, focus on simplifying the mathematical expression in the right-hand side of the equation.

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

The limit of the sum of two functions can be evaluated by the sum of their limits as per sum rule of limits.

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

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

It can be further simplified by using product rule of limits.

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

### Evaluate Limits of the functions

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

Find the limit of first factor of each term in the right-hand side of the equation by the direct substitution method.

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

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

Observe the second factor in each term of the expression. The second factors in both terms represent the differentiation of functions $f{(x)}$ and $g{(x)}$ as per definition of the derivative in limiting operation.

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

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

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

According to the derivative of function by first principle, the differentiation of product of two functions is equal to sum of product of first function and derivative of second function, and the product of second function and derivative of first function.

It is called as the product rule of differentiation in differential calculus. The derivative product rule is also written in terms of $u$ and $v$ by taking $u = f{(x)}$ and $v = g{(x)}$ in calculus.

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, (u.v)}$ $\,=\,$ $u\dfrac{dv}{dx}$ $+$ $v\dfrac{du}{dx}$

Sometimes, the product of derivative is sometimes called as $uv$ rule by some people.

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