Question

What is the derivative of $\sqrt{2x}$?

Answer 1

Write the given radical expression into the exponent form with exponent equal to $\dfrac{1}{2}$. Now, use the property of exponents given as ${{\left( a\times b \right)}^{m}}={{a}^{m}}\times {{b}^{m}}$ to remove the constant term from the coefficient of the variable x. Differentiate the function and use the formula $\dfrac{d\left[ {{x}^{n}} \right]}{dx}=n{{x}^{n-1}}$ to get the required derivative. Use the fact that if a constant is multiplied to a function then we can take out that constant from the derivative formula.

Complete step-by-step solution:
Here we have been provided with the function $\sqrt{2x}$ and we are asked to find its derivative. Assuming the given expression function as y we have,
$\Rightarrow y=\sqrt{2x}$
Converting the radical form of the expression into the exponential form we get,
$\Rightarrow y={{\left( 2x \right)}^{\dfrac{1}{2}}}$
Using the property of exponents given as ${{\left( a\times b \right)}^{m}}={{a}^{m}}\times {{b}^{m}}$ we get,
$\Rightarrow y={{\left( 2 \right)}^{\dfrac{1}{2}}}\times {{\left( x \right)}^{\dfrac{1}{2}}}$
On differentiating both the sides with respect to x we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( {{\left( 2 \right)}^{\dfrac{1}{2}}}\times {{\left( x \right)}^{\dfrac{1}{2}}} \right)}{dx}$
Clearly we can see that ${{2}^{\dfrac{1}{2}}}$ is a constant so it can be directly taken out of the derivative, therefore we get,
$\Rightarrow \dfrac{dy}{dx}={{\left( 2 \right)}^{\dfrac{1}{2}}}\times \dfrac{d\left( {{\left( x \right)}^{\dfrac{1}{2}}} \right)}{dx}$
Using the formula $\dfrac{d\left[ {{x}^{n}} \right]}{dx}=n{{x}^{n-1}}$ we get,

& \Rightarrow \dfrac{dy}{dx}={{\left( 2 \right)}^{\dfrac{1}{2}}}\times \dfrac{1}{2}{{\left( x \right)}^{\dfrac{1}{2}-1}} \\\ & \Rightarrow \dfrac{dy}{dx}={{\left( 2 \right)}^{\dfrac{1}{2}}}\times \dfrac{1}{2}{{\left( x \right)}^{-\dfrac{1}{2}}} \\\ \end{aligned}$$ Converting this exponent form back into the radical form we get, $$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=\sqrt{2}\times \dfrac{1}{2\sqrt{x}} \\\ & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{2}\times \sqrt{x}} \\\ & \therefore \dfrac{dy}{dx}=\dfrac{1}{\sqrt{2x}} \\\ \end{aligned}$$ Hence, the above relation is our answer. **Note:** You must remember all the basic rules and formulas of differentiation like: - product rule, chain rule, $$\dfrac{u}{v}$$ rule etc. From the above solution you may note an important formula given as $$\dfrac{d\left[ \sqrt{f\left( x \right)} \right]}{dx}=\dfrac{1}{2\sqrt{f\left( x \right)}}\times \dfrac{d\left[ f\left( x \right) \right]}{dx}$$. It is an important formula we directly use in other chapters of mathematics. That fact that ‘the constant can be taken out of the derivative’ arises from the theorem that the derivative of a constant is 0, so applying the product rule of derivative we get the above conclusion.