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Proof of $\dfrac{d}{dx} \sqrt{x}$ in Limit Method


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


The differentiation of function $\sqrt{x}$ with respect to $x$ can be derived in differential calculus by using limit method. It is a fundamental method of deriving derivative of any function including $\sqrt{x}$.

Differentiation of function in Limit method

The derivative of a function can be derived fundamentally by the expression of differentiation of function in limit form.

$\dfrac{d}{dx} f{(x)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0} \normalsize \dfrac{f{(x+h)}-f{(x)}}{h}$

Take $f{(x)} \,=\, \sqrt{x}$, then $f{(x+h)} \,=\, \sqrt{x+h}$. Now, substitute them in the mathematical relation to start deriving the differentiation of the function $\sqrt{x}$ with respect to $x$ in differential calculus.

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

Now, substitute $h \,=\, 0$ to find the value of the function as the limit $h$ approaches zero.

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

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

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

The value of the function is indeterminate as the limit $h$ tends to $0$ but the derivative of $\sqrt{x}$ with respect to $x$ cannot be indeterminate. So, an alternate mathematical approach should be used to differentiate the function $\sqrt{x}$.

Try Rationalization method

The mathematical function is in radical form. So, use rationalising method to multiply the expression in numerator by its conjugate function.

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

Simplify the mathematical expression

Now, multiply the factors in both numerator and denominator, and then simply the mathematical expression to move ahead in deriving the derivative of $\sqrt{x}$ with respect to $x$ in calculus.

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

The numerator represents product of sum and difference of two terms. It can be simplified by (a+b)(a-b) formula.

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

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

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

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

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

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

Evaluate the function

Replace $h$ by zero to find the value of the function as the limit $h$ approaches $0$.

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

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

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

Therefore, it is proved that the derivative of square root of $x$ with respect to $x$ is equal to reciprocal of twice the square root of $x$. It is used as formula in differential calculus.

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