Question

How do you differentiate ${\left( {1 + \left( {\dfrac{1}{x}} \right)} \right)^3}$ ?

Answer 1

Hint : Here we need to differentiate the given problem with respect to x. here we use the algebraic identity ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$ and then we differentiate it. We know that the differentiation of constant term is zero and differentiation of ${x^n}$ is $\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}$ .

Complete step-by-step answer :
Given,
${\left( {1 + \left( {\dfrac{1}{x}} \right)} \right)^3}$ .
Now applying ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$ we have,
${\left( {1 + \left( {\dfrac{1}{x}} \right)} \right)^3} = {1^3} + {\left( {\dfrac{1}{x}} \right)^3} + 3\left( 1 \right)\left( {\dfrac{1}{x}} \right)\left( {1 + \dfrac{1}{x}} \right)$
$= 1 + \dfrac{1}{{{x^3}}} + \dfrac{3}{x}\left( {1 + \dfrac{1}{x}} \right)$
$= 1 + \dfrac{1}{{{x^3}}} + \dfrac{3}{x} + \dfrac{3}{{{x^2}}}$
$= 1 + {x^{ - 3}} + 3{x^{ - 1}} + 3{x^{ - 2}}$
Thus we have
${\left( {1 + \left( {\dfrac{1}{x}} \right)} \right)^3} = 1 + {x^{ - 3}} + 3{x^{ - 1}} + 3{x^{ - 2}}$
Now differentiating with respect to ‘x’,
$\dfrac{d}{{dx}}{\left( {1 + \left( {\dfrac{1}{x}} \right)} \right)^3} = \dfrac{d}{{dx}}\left( {1 + {x^{ - 3}} + 3{x^{ - 1}} + 3{x^{ - 2}}} \right)$
By linear combination rule,
$\dfrac{d}{{dx}}{\left( {1 + \left( {\dfrac{1}{x}} \right)} \right)^3} = \dfrac{d}{{dx}}\left( 1 \right) + \dfrac{d}{{dx}}\left( {{x^{ - 3}}} \right) + 3\dfrac{d}{{dx}}\left( {{x^{ - 1}}} \right) + 3\dfrac{d}{{dx}}\left( {{x^{ - 2}}} \right)$
Differentiation of constant is zero,
$\dfrac{d}{{dx}}{\left( {1 + \left( {\dfrac{1}{x}} \right)} \right)^3} = 0 + - 3{x^{ - 3 - 1}} + 3\left( { - 1.{x^{ - 1 - 1}}} \right) + 3\left( { - 2.{x^{ - 2 - 1}}} \right)$
$\dfrac{d}{{dx}}{\left( {1 + \left( {\dfrac{1}{x}} \right)} \right)^3} = - 3{x^{ - 4}} + 3\left( { - 1.{x^{ - 2}}} \right) + 3\left( { - 2.{x^{ - 3}}} \right)$
$\dfrac{d}{{dx}}{\left( {1 + \left( {\dfrac{1}{x}} \right)} \right)^3} = - 3{x^{ - 4}} - 3{x^{ - 2}} - 6.{x^{ - 3}}$ . This is the required result.
So, the correct answer is “$- 3{x^{ - 4}} - 3{x^{ - 2}} - 6.{x^{ - 3}}$”.

Note : $\bullet$ Linear combination rule: The linearity law is very important to emphasize its nature with alternate notation. Symbolically it is specified as $h'(x) = af'(x) + bg'(x)$
$\bullet$ Quotient rule: The derivative of one function divided by other is found by quotient rule such as ${\left[ {\dfrac{{f(x)}}{{g(x)}}} \right] ^1} = \dfrac{{g(x)f'(x) - f(x)g'(x)}}{{{{\left[ {g(x)} \right] }^2}}}$ .
$\bullet$ Product rule: When a derivative of a product of two function is to be found, then we use product rule that is $\dfrac{{dy}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}$ .
$\bullet$ Chain rule: To find the derivative of composition function or function of a function, we use chain rule. That is $fog'({x_0}) = [(f'og)({x_0})] g'({x_0})$ .