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.