$x$ is a variable and its square root is written as $\sqrt{x}$. The derivative of square root of $x$ with respect to $x$ is written in differential calculus as follows.

$\dfrac{d}{dx}{(\sqrt{x})}$

According to definition of the derivative, the differentiation of $\sqrt{x}$ with respect to $x$ can be written in limiting operation form.

$\dfrac{d}{dx}{\sqrt{x}}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to \, 0}{\normalsize \dfrac{\sqrt{x+\Delta x}-\sqrt{x}}{\Delta x}}$

Now, take $\Delta x = h$ and convert the equation in terms of $h$ from $\Delta x$.

$\implies$ $\dfrac{d}{dx}{\sqrt{x}}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize \dfrac{\sqrt{x+h}-\sqrt{x}}{h}}$

Now, let’s calculate the differentiation of square root of $x$ with respect to $x$ by first principle.

The limit of the algebraic function in radical form can be evaluated by the direct substitution method.

$\implies$ $\dfrac{d}{dx}{\sqrt{x}}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize \dfrac{\sqrt{x+h}-\sqrt{x}}{h}}$

$=\,\,\,$ $\dfrac{\sqrt{x+0}-\sqrt{x}}{0}$

$=\,\,\,$ $\dfrac{\sqrt{x}-\sqrt{x}}{0}$

$=\,\,\,$ $\dfrac{0}{0}$

It is calculated that the limit of the algebraic function is indeterminate. So, it is not possible to find the derivative of square root of $x$ by direct substation method of limits. So, it should be calculated in another method.

The function is in radical form. So, let us try to evaluate it by rationalisation method by rationalizing the function in numerator by its conjugate function.

$\dfrac{d}{dx}{\sqrt{x}}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize \dfrac{\sqrt{x+h}-\sqrt{x}}{h}}$ $\times$ $1$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize \dfrac{\sqrt{x+h}-\sqrt{x}}{h}}$ $\times$ $\dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize \dfrac{{(\sqrt{x+h}-\sqrt{x})}{(\sqrt{x+h}+\sqrt{x})}}{h \times (\sqrt{x+h}+\sqrt{x})}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize \dfrac{{(\sqrt{x+h})}^2-{(\sqrt{x})}^2}{h \times (\sqrt{x+h}+\sqrt{x})}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize \dfrac{x+h-x}{h \times (\sqrt{x+h}+\sqrt{x})}}$

$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{h \,\to \, 0}{\normalsize \dfrac{\cancel{x}+h-\cancel{x}}{h \times (\sqrt{x+h}+\sqrt{x})}}$

$=\,\,\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize \dfrac{h}{h \times (\sqrt{x+h}+\sqrt{x})}}$

$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{h \,\to \, 0}{\normalsize \dfrac{\cancel{h}}{\cancel{h} \times (\sqrt{x+h}+\sqrt{x})}}$

$\implies$ $\dfrac{d}{dx}{\sqrt{x}}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to \, 0}{\normalsize \dfrac{1}{\sqrt{x+h}+\sqrt{x}}}$

Now, evaluate the limit of the algebraic function as $h$ approaches $0$ by direct substitution method.

$\implies$ $\dfrac{d}{dx}{\sqrt{x}}$ $\,=\,$ $\dfrac{1}{\sqrt{x+0}+\sqrt{x}}$

$\implies$ $\dfrac{d}{dx}{\sqrt{x}}$ $\,=\,$ $\dfrac{1}{\sqrt{x}+\sqrt{x}}$

$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\sqrt{x}}$ $\,=\,$ $\dfrac{1}{2\sqrt{x}}$

Therefore, it is proved from first principle that the derivative of square root of $x$ is equal to the quotient of $1$ by two times square root of $x$.

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