# Derivative of x root(x) by First Principle

The derivative of x root x is 3 root(x)/2. In this post, we will find the derivative of x root(x) from first principle. This principle says that the derivative of f(x) is

$\dfrac{d}{dx}(f(x))$ $=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$ $\quad \cdots (I)$

## Derivative of x root x by First Principle

Let $f(x)=x\sqrt{x}$ in the above formula (I). So the derivative of x root x using first principle is equal to

$\dfrac{d}{dx}(x \sqrt{x})$ $=\lim\limits_{h \to 0}\dfrac{(x+h)\sqrt{x+h}-x\sqrt{x}}{h}$

We will multiply the numerator with its conjugate.

$=\lim\limits_{h \to 0}[\dfrac{(x+h)\sqrt{x+h}-x\sqrt{x}}{h}$ $\times \dfrac{(x+h)\sqrt{x+h}+x\sqrt{x}}{(x+h)\sqrt{x+h}+x\sqrt{x}}]$

Apply the formula of $(a-b)(a+b)$ $=a^2-b^2$ in the numerator. By doing that we get

$=\lim\limits_{h \to 0}\dfrac{[(x+h)\sqrt{x+h}]^2-(x\sqrt{x})^2}{h((x+h)\sqrt{x+h}+x\sqrt{x})}$

$=\lim\limits_{h \to 0}\dfrac{(x+h)^3-x^3}{h((x+h)\sqrt{x+h}+x\sqrt{x})}$

Now, we will use the formula of $a^3-b^3$ $=(a-b)(a^2+ab+b^2)$ in the numerator. So the above limit

$=\lim\limits_{h \to 0}\dfrac{h(3x^2+3xh+h^2)}{h((x+h)\sqrt{x+h}+x\sqrt{x})}$

$=\lim\limits_{h \to 0}\dfrac{3x^2+3xh+h^2}{(x+h)\sqrt{x+h}+x\sqrt{x}}$

$=\dfrac{3x^2+3x \cdot 0+0^2}{(x+0)\sqrt{x+0}+x\sqrt{x}}$ $=\dfrac{3x^2}{2x\sqrt{x}}$

$=\dfrac{3\sqrt{x}}{2}$

Thus, the derivative of x root(x) is 3 root(x)/2 obtained by the first principle of derivatives.

Note: The derivative of r root(x) can be computed by the power rule of derivatives easily. See that

$x\sqrt{x}=x^{1+1/2}$ $=x^{3/2}$.

$\therefore \dfrac{d}{dx}(x\sqrt{x})$ $=\dfrac{d}{dx}(x^{\frac{3}{2}})$

$=3/2 x^{3/2-1}$ as we know that $\dfrac{d}{dx}(x^n)=nx^{n-1}$. Here $n=3/2$.

$=3/2 x^{1/2}$

$=\dfrac{3\sqrt{x}}{2}$

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## FAQs

Q1: What is the derivative of x√x?

Answer: The derivative of x√x is equal to 3√x/2.