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.