# Proof for Reciprocal rule of Derivatives

The derivative of the reciprocal of a function is equal to the negative product of $1$ by square of the function and the derivative of the function.

$\dfrac{d}{dx}{\bigg(\dfrac{1}{f(x)}\bigg)}$ $\,=\,$ $-\,\dfrac{1}{\Big(f(x)\Big)^2}$ $\times$ $\dfrac{d}{dx}{\Big(f(x)\Big)}$

Let’s learn how to prove the reciprocal rule of the derivatives mathematically.

### Definition of the Derivative in Limit form

According to the definition of the derivatives, the definition of the derivative of functions $f(x)$ and $g(x)$ are expressed in limit form.

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

$(2).\,\,\,$ $\dfrac{d}{dx}{\Big(g(x)\Big)}$ $\,=\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{g(x+h)-g(x)}{h}}$

### Differentiation of the Reciprocal function

Let’s consider the differentiation from the first principle, the derivative of the function $g(x)$ with respect to $x$ is written as follows.

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

Now, let us assume that the reciprocal of the function $f(x)$ is $g(x)$.

$g(x) \,=\, \dfrac{1}{f(x)}$

Then,
$g(x+h) \,=\, \dfrac{1}{f(x+h)}$

Now, substitute them in the definition of the differentiation of the function $g(x)$.

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

Let us concentrate on simplifying the right hand side expression of the equation.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\dfrac{1}{f(x+h)}-\dfrac{1}{f(x)}}{h}}$

Use the subtraction rule of the fractions in the numerator to find their difference.

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\dfrac{1 \times f(x)- f(x+h) \times 1}{f(x+h) \times f(x)}}{h}}$

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

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

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\dfrac{(-1) \times \Big(f(x+h)-f(x)\Big)}{f(x+h) \times f(x)}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{(-1) \times \dfrac{f(x+h)-f(x)}{f(x+h) \times f(x)}}{h}}$

$=\,\,\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg((-1) \times \dfrac{\dfrac{f(x+h)-f(x)}{f(x+h) \times f(x)}}{h}\Bigg)}$

Now, release the constant factor from the limit operation by using the constant multiple rule of the limits.

$=\,\,\,$ $(-1) \times \displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\dfrac{f(x+h)-f(x)}{f(x+h) \times f(x)}}{h}}$

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

$=\,\,\,$ $-\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\dfrac{\Big(f(x+h)-f(x)\Big) \times 1}{f(x+h) \times f(x)}}{h}}$

For our convenience, split the fraction into product of two multiplying factors.

$=\,\,\,$ $-\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{\Big(f(x+h)-f(x)\Big) \times \dfrac{1}{f(x+h) \times f(x)}}{h}}$

Once again, split the fraction into product of two multiplying factors.

$=\,\,\,$ $-\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(\dfrac{f(x+h)-f(x)}{h}}$ $\times$ $\dfrac{1}{f(x+h) \times f(x)}\Bigg)$

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{d}{dx}{\bigg(\dfrac{1}{f(x)}\bigg)}$ $\,=\,$ $-\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(\dfrac{f(x+h)-f(x)}{h}}$ $\times$ $\dfrac{1}{f(x+h) \times f(x)}\Bigg)$

### Evaluate the Limits of the functions

According to the product rule of the limits, the limit of the product of two functions can be evaluated by the product of their limits.

$=\,\,\,$ $-\Bigg(\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{f(x+h)-f(x)}{h}}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{1}{f(x+h) \times f(x)}\Bigg)}$

According to the commutative property of multiplication, the product of the factors can be written as follows.

$=\,\,\,$ $-\Bigg(\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{1}{f(x+h) \times f(x)}}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{f(x+h)-f(x)}{h}\Bigg)}$

The limit of the first factor can be calculated by the direct substitution method.

$=\,\,\,$ $-\Bigg(\dfrac{1}{f(x+0) \times f(x)}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{f(x+h)-f(x)}{h}\Bigg)}$

$=\,\,\,$ $-\Bigg(\dfrac{1}{f(x) \times f(x)}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{f(x+h)-f(x)}{h}\Bigg)}$

$=\,\,\,$ $-\Bigg(\dfrac{1}{\Big(f(x)\Big)^2}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{f(x+h)-f(x)}{h}\Bigg)}$

$=\,\,\,$ $-\,\dfrac{1}{\Big(f(x)\Big)^2}$ $\times$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{f(x+h)-f(x)}{h}}$

The second factor is the fundamental definition of the derivative of the function $f(x)$ with respect to $x$ in limit form.

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{d}{dx}{\bigg(\dfrac{1}{f(x)}\bigg)}$ $\,=\,$ $-\,\dfrac{1}{\Big(f(x)\Big)^2}$ $\times$ $\dfrac{d}{dx}{\Big(f(x)\Big)}$

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