Differentiate the function $\left( {\dfrac{x}{{x + 1}}} \right)$ using the first principle of differentiation?

Hint: The given question wants us to evaluate the derivative of a given rational function $\left( {\dfrac{x}{{x + 1}}} \right)$ using the first principle of derivative. The first principle of differentiation involves the concepts of limits, derivatives, continuity and differentiability. It helps us to determine the derivative of the function using concepts of limits.

Complete step-by-step answer:
Given the function $\left( {\dfrac{x}{{x + 1}}} \right)$, we have to calculate the derivative of the function using the first principle of derivatives.
Using formula of first principle of derivatives, we know that derivative of a function is calculated as:
\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}\]
So, derivative of the given function$f(x) = \cos ec\left( x \right)$,
$ \Rightarrow $\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{{\left( {x + h} \right)}}{{\left( {x + h} \right) + 1}} - \dfrac{x}{{x + 1}}}}{h}\]
Taking LCM in the numerator, we get,
$ \Rightarrow $\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\dfrac{{\left( {x + h} \right)\left( {x + 1} \right) - x\left( {\left( {x + h} \right) + 1} \right)}}{{\left( {\left( {x + h} \right) + 1} \right)\left( {x + 1} \right)}}}}{h}\]
Multiplying the terms in numerator, we get,
$ \Rightarrow $\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{x^2} + hx + x + h - {x^2} - xh - x}}{{\left( {\left( {x + h} \right) + 1} \right)\left( {x + 1} \right)h}}\]
Cancelling like terms with opposite signs, we get,
$ \Rightarrow $\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{h}{{\left( {\left( {x + h} \right) + 1} \right)\left( {x + 1} \right)h}}\]
Cancelling common factors in numerator and denominator, we get,
$ \Rightarrow $\[f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{{\left( {\left( {x + h} \right) + 1} \right)\left( {x + 1} \right)}}\]
Now, putting in the value of h in the limit, we get,
$ \Rightarrow $\[f'(x) = \dfrac{1}{{\left( {\left( {x + 0} \right) + 1} \right)\left( {x + 1} \right)}}\]
$ \Rightarrow $\[f'(x) = \dfrac{1}{{\left( {x + 1} \right)\left( {x + 1} \right)}}\]
Simplifying the expression further, we get,
$ \Rightarrow $\[f'(x) = \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}\]
So, we get the derivative of $\left( {\dfrac{x}{{x + 1}}} \right)$ as \[\dfrac{1}{{{{\left( {x + 1} \right)}^2}}}\] using the first principle of differentiation.

Note: There are various methods of finding the derivatives of the given functions.
The given function can also be written as $\left( {\dfrac{x}{{x + 1}}} \right) = \left( {\dfrac{{x + 1 - 1}}{{x + 1}}} \right) = 1 - \left( {\dfrac{1}{{x + 1}}} \right)$.
Now, when we differentiate it with respect to x using the chain rule of differentiation $\dfrac{{d\left( {f\left( {g\left( x \right)} \right)} \right)}}{{dx}} = f'\left( {g\left( x \right)} \right) \times g'\left( x \right)$, we get,
$f'\left( x \right) = \dfrac{d}{{dx}}\left[ {1 - \left( {\dfrac{1}{{x + 1}}} \right)} \right] = - \dfrac{d}{{dx}}\left( {\dfrac{1}{{x + 1}}} \right)$
$ \Rightarrow f'\left( x \right) = - \left( { - 1} \right)\left( {\dfrac{1}{{{{\left( {x + 1} \right)}^2}}}} \right) = \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}$
So, we get the same answer from both the methods.