Math Doubts

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.

Math Questions

The math problems with solutions to learn how to solve a problem.

Learn solutions

Math Worksheets

The math worksheets with answers for your practice with examples.

Practice now

Math Videos

The math videos tutorials with visual graphics to learn every concept.

Watch now

Subscribe us

Get the latest math updates from the Math Doubts by subscribing us.

Learn more

Math Doubts

A free math education service for students to learn every math concept easily, for teachers to teach mathematics understandably and for mathematicians to share their maths researching projects.

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved