Question

What is the derivative of $\sqrt{{{x}^{3}}}$?

Answer 1

We try to form the indices formula for the value 2. This is a square root of ${{x}^{3}}$. We take the indices form of ${{x}^{3}}$. We multiply the fraction with 3 to find the simplified form. Then we use the formula of $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ to find the derivative of $\sqrt{{{x}^{3}}}$.

Complete step by step solution:
We need to find the value of the algebraic form of $\sqrt{{{x}^{3}}}$. This is a square root form.
The given value is the form of indices. We are trying to find the root value of ${{x}^{3}}$.
We know the theorem of indices ${{a}^{\dfrac{1}{n}}}=\sqrt[n]{a}$. Putting value 2 we get ${{a}^{\dfrac{1}{2}}}=\sqrt[2]{a}$.
Therefore, $\sqrt{{{x}^{3}}}={{\left( {{x}^{3}} \right)}^{\dfrac{1}{2}}}$. We know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$.
So, we get $\sqrt{{{x}^{3}}}={{\left( {{x}^{3}} \right)}^{\dfrac{1}{2}}}={{x}^{\dfrac{3}{2}}}$ .
Now we find the derivative of ${{x}^{\dfrac{3}{2}}}$.
We use the derivative formula of $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$.
We find $\dfrac{d}{dx}\left( {{x}^{\dfrac{3}{2}}} \right)=\dfrac{3}{2}{{x}^{\dfrac{3}{2}-1}}=\dfrac{3}{2}{{x}^{\dfrac{1}{2}}}=\dfrac{3\sqrt{x}}{2}$.
Therefore, the derivative of $\sqrt{{{x}^{3}}}$ is $\dfrac{3\sqrt{x}}{2}$.

Note:
The derivative form of $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ is applicable for all values of n\in \mathbb{R} - \left\\{ 0 \right\\}. We also could have used chain rule where we need remember that in the chain rule $\dfrac{d}{d\left[ h\left( x \right) \right]}\left[ goh\left( x \right) \right]\times \dfrac{d\left[ h\left( x \right) \right]}{dx}$, we aren’t cancelling out the part $d\left[ h\left( x \right) \right]$.