Question

How do you solve the derivative of \[\sqrt {2x} \] ?

Answer 1

Hint: Here we need to differentiate the given problem with respect to x. We know that the differentiation of \[{x^n}\] with respect to ‘x’ is \[\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}\] . We know that \[\sqrt x \] means that x to the power \[\dfrac{1}{2}\], that is \[{x^{\dfrac{1}{2}}}\] . We use this concept to solve the given problem.

Complete step by step solution:
Given,
 \[\sqrt {2x} \] .
That is we have \[\sqrt {2x} = {\left( {2x} \right)^{\dfrac{1}{2}}}\] .
Now differentiating this with respect to ‘x,
 \[\dfrac{d}{{dx}}\left( {\sqrt {2x} } \right) = \dfrac{d}{{dx}}\left( {{{\left( {2x} \right)}^{\dfrac{1}{2}}}} \right)\]
We know that the differentiation of \[{x^n}\] with respect to ‘x’ is \[\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}\] ,
 \[ = \dfrac{1}{2}{\left( {2x} \right)^{\dfrac{1}{2} - 1}}.\dfrac{d}{{dx}}\left( {2x} \right)\]
 \[ = \dfrac{2}{2}{\left( {2x} \right)^{ - \dfrac{1}{2}}}.\dfrac{d}{{dx}}\left( x \right)\]
 \[ = {\left( {2x} \right)^{ - \dfrac{1}{2}}}\]
Because differentiation of ‘x’ with respect to x is one.
Thus we have,
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {\sqrt {2x} } \right) = {\left( {2x} \right)^{ - \dfrac{1}{2}}}\]
Or
 \[ \Rightarrow \dfrac{d}{{dx}}\left( {\sqrt {2x} } \right) = \dfrac{1}{{\sqrt {2x} }}\] . This is the required result.
So, the correct answer is “ \[\dfrac{1}{{\sqrt {2x} }}\] ”.

Note: We know the differentiation of \[{x^n}\] is \[\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}\] . The obtained result is the first derivative. If we differentiate again we get a second derivative. If we differentiate the second derivative again we get a third derivative and so on. Careful in applying the product rule. We also know that differentiation of constant terms is zero.