Question

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

Answer 1

Hint: 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 \[d\left[ h\left( x \right) \right]\].