# Derivative of root(x)+1/root(x)

Here, we will find the derivative of root x+1/root x with respect to x. In the end, we will also evaluate this derivative at x=1.

For more details of square roots, please click on the page Square Root of x: Definition, Symbol, Graph, Properties, Derivative, Integration

## Root(x)+1/Root(x) Derivative

Question: Find the derivative of $\sqrt{x}+\frac{1}{\sqrt{x}}$

At first, we will rewrite the given quantity $\sqrt{x}+\frac{1}{\sqrt{x}}$ using the rule of exponents. So we have

$\sqrt{x}+\frac{1}{\sqrt{x}}$

$=x^{1/2}+\frac{1}{x^{1/2}}$

$=x^{1/2}+x^{-1/2}$ …(i)

Thus, the desired derivative is

$=\dfrac{d}{dx}(\sqrt{x}+\frac{1}{\sqrt{x}})$

$=\dfrac{d}{dx}(x^{\frac{1}{2}}+x^{-\frac{1}{2}})$ by (i)

$=\dfrac{d}{dx}(x^{\frac{1}{2}})+\frac{d}{dx}(x^{-\frac{1}{2}})$

$=\dfrac{1}{2} x^{\frac{1}{2} -1} -\frac{1}{2} x^{-\frac{1}{2} -1}$

[by the power rule of derivatives $\frac{d}{dx}(x^n) = nx^{n-1}$]

$=\dfrac{1}{2} x^{-\frac{1}{2}} -\frac{1}{2} x^{-\frac{3}{2}}$

$=\dfrac{1}{2} [\frac{1}{x^{\frac{1}{2}}} – \frac{1}{x^\frac{3}{2}}]$

$=\dfrac{1}{2} [\frac{1}{x^{\frac{1}{2}}} -\frac{1}{x.x^{\frac{1}{2}}}]$

$=\dfrac{1}{2} [\frac{1}{\sqrt{x}} -\frac{1}{x\sqrt{x}}]$

$=\frac{1}{2\sqrt{x}} [1 -\frac{1}{{x}}]$

So the derivative of rootx+1/rootx is $=\frac{1}{2\sqrt{x}} [1 -\frac{1}{{x}}]$, that is,

d/dx(√x+1/√x) = (1/2√x) [1 -1/x].

Integration of Root x

Question: Find the derivative of $\sqrt{x}+\frac{1}{\sqrt{x}}$ at $x=1$

We have obtained the derivative of root x +1/root x above. So putting $x=1$ in the above derivative, we will get the answer.

$\dfrac{d}{dx}[\sqrt{x}+\frac{1}{\sqrt{x}}]$ at $x=1$

$=\frac{1}{2\sqrt{x}} [1 -\frac{1}{x}]$ at x=1

$=\frac{1}{2\sqrt{1}} [1 -\frac{1}{1}]$

$=\dfrac{1}{2}[1-1]$

$=0$

## Derivative of root x -1

The derivative of root(x)-1 is equal to

$\dfrac{d}{dx}(\sqrt{x}-1)$ $=\dfrac{d}{dx}(\sqrt{x}) – \dfrac{d}{dx}(1)$

$=\dfrac{1}{2} x^{\frac{1}{2} -1}-0$ as the derivative of a constant is zero.

$=\dfrac{1}{2}x^{-\frac{1}{2}}$

$=\frac{1}{2\sqrt{x}}$

Related Keywords:

√x+1/√x differentiate | root x + 1/root x differentiation | derivative of root x + 1 by root x

## FAQs

Q1: Find the derivative of rootx+1/rootx.

Answer: The derivative of rootx+1/rootx is equal to $=\frac{1}{2\sqrt{x}} [1 -\frac{1}{{x}}]$.