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 $${\displaystyle \frac{d(x^n)}{dx}}=n.x^{n-1}$$ . We know that $$\sqrt{x}$$ means that x to the power $${\displaystyle \frac{1}{2}}$$, that is $$x^{{\displaystyle \frac{1}{2}}}$$ . We use this concept to solve the given problem.

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

Note: We know the differentiation of $$x^n$$ is $${\displaystyle \frac{d(x^n)}{dx}}=n.x^{n-1}$$ . 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.