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.
According to the definition of the derivatives, the definition of the derivative of functions $f(x)$ and $g(x)$ are expressed in limit form.
$(1).\,\,\,$ $\dfrac{d}{dx}{\Big(f(x)\Big)}$ $\,=\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{f(x+h)-f(x)}{h}}$
$(2).\,\,\,$ $\dfrac{d}{dx}{\Big(g(x)\Big)}$ $\,=\,$ $\displaystyle \large \lim_{h\,\to\,0}{\normalsize \dfrac{g(x+h)-g(x)}{h}}$
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.
$=\,\,\,$ $-\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(\dfrac{f(x+h)-f(x)}{h}}$ $\times$ $\dfrac{1}{f(x+h) \times f(x)}\Bigg)$
$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{d}{dx}{\bigg(\dfrac{1}{f(x)}\bigg)}$ $\,=\,$ $-\displaystyle \large \lim_{h\,\to\,0}{\normalsize \Bigg(\dfrac{f(x+h)-f(x)}{h}}$ $\times$ $\dfrac{1}{f(x+h) \times f(x)}\Bigg)$
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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