Question

Derive the expression : $\sqrt 2 x + 2\sqrt x - \dfrac{1}{{\sqrt x }}$.
(a) $\sqrt 2 + \dfrac{1}{{\sqrt x }}\left( {1 + \dfrac{1}{{2x}}} \right)$
(b) $\sqrt 2 + \dfrac{1}{{\sqrt x }}\left( {1 - \dfrac{1}{{2x}}} \right)$
(c) Cannot be determined
(d) None of the above

Answer 1

Hint : We will use the most eccentric concept of derivations. Considering the ‘y’ variable as a derivating agent or term the solution is solved by using laws of derivation such as $\dfrac{d}{{dx}}{\left( x \right)^n} = n{x^{n - 1}}$, $\dfrac{d}{{dx}}\left( {\sqrt x } \right) = \dfrac{1}{{2\sqrt x }}$ , $\dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right) = - \dfrac{1}{{{x^2}}}$ . As a result, substituting the values in the given expression one can easily solve the complete problem. Keenly solve the problem which might get confused in a wise manner!

Complete step-by-step answer :
Since, we have the given expression
$\sqrt 2 x + 2\sqrt x - \dfrac{1}{{\sqrt x }}$
As a result, solving the given expression, first of all derivating the above given equation with respect to the ‘x’ variable, can reach up to a desire output,
Hence, derivating each term individually, we get
($1$) $\sqrt 2 x = \dfrac{d}{{dx}}\left( {\sqrt 2 x} \right)$
Where, $\sqrt 2$is constant,
$\sqrt 2 x = \sqrt 2 \dfrac{d}{{dx}}\left( x \right) = \sqrt 2 x = \sqrt 2 \dfrac{d}{{dx}}{\left( x \right)^1}$
Using the rule of derivative/s that is$\dfrac{d}{{dx}}{\left( x \right)^n} = n{x^{n - 1}}$, we get
$\sqrt 2 x = \dfrac{d}{{dx}}\left( {\sqrt 2 x} \right) = \sqrt 2$ … (i)
Similarly,
($2$)$2\sqrt x = \dfrac{d}{{dx}}\left( {2\sqrt x } \right)$
Here, $2$is constant,
$2\sqrt x = 2\dfrac{d}{{dx}}\sqrt x$
Using the rule of derivative/s that is $\dfrac{d}{{dx}}\left( {\sqrt x } \right) = \dfrac{1}{{2\sqrt x }}$ , we get

2\sqrt x = \dfrac{d}{{dx}}\left( {2\sqrt x } \right) = 2 \times \dfrac{1}{{2\sqrt x }} \\\ 2\sqrt x = \dfrac{d}{{dx}}\left( {2\sqrt x } \right) = \dfrac{1}{{\sqrt x }} \\\ $$ … (ii) And, ($3$) $\dfrac{1}{{\sqrt x }} = \dfrac{d}{{dx}}\left( {\dfrac{1}{{\sqrt x }}} \right)$ Using the rule of derivative/s that is, $\dfrac{d}{{dx}}\left( {\dfrac{1}{{\sqrt x }}} \right) = - \dfrac{1}{x} \times \dfrac{1}{{2\sqrt x }}$ Here, first of all taking the derivative of $\dfrac{1}{x}$and then using the taking the derivative of $$\sqrt x $$ , we get $$\dfrac{d}{{dx}}\left( {\dfrac{1}{{\sqrt x }}} \right) = - \dfrac{1}{{2x\sqrt x }}$$ $$\therefore \dfrac{1}{{\sqrt x }} = \dfrac{d}{{dx}}\left( {\dfrac{1}{{\sqrt x }}} \right) = - \dfrac{1}{{2x\sqrt x }}$$ … (iii) Now, from (i), (ii) and (iii), Substituting all the values in the given expression, we get

\sqrt 2 x + 2\sqrt x - \dfrac{1}{x} = \sqrt 2 + \dfrac{1}{{\sqrt x }} - \left( { - \dfrac{1}{{2x\sqrt x }}} \right) \\
\sqrt 2 x + 2\sqrt x - \dfrac{1}{x} = \sqrt 2 + \dfrac{1}{{\sqrt x }} + \dfrac{1}{{2x\sqrt x }} \\

Solving the equation mathematically, we get $$\sqrt 2 x + 2\sqrt x - \dfrac{1}{x} = \sqrt 2 + \dfrac{1}{{\sqrt x }}\left( {1 + \dfrac{1}{{2x}}} \right)$$ Therefore, the option (a) is correct! **So, the correct answer is “Option a”.** **Note** : One must remember the concept of derivation, how to differentiate the equation with respect to which variable, etc.? Also, laws of derivation such as $\dfrac{d}{{dx}}(xy) = x\dfrac{{dy}}{{dx}} + y(1)$is the multiplication rule of derivation used here and $\dfrac{{dy}}{{dx}},\dfrac{{{d^2}y}}{{d{x^2}}},...$ can be noted as and so on! Derivation of any constant number is always zero. Deriving the equation with the same term is always one$\dfrac{d}{{dx}}(x) = 1$. Algebraic identities play a significant role in solving this problem.