Question

Differentiate the following with repect to x

1). (root x + 1/root x)

Answer 1

Solution:

$\sf{\implies \Bigg(\sqrt{x} + \dfrac{1}{\sqrt{x}}\Bigg)}$

$\sf{\implies \dfrac{d\Bigg(\sqrt{x}+\dfrac{1}{\sqrt{x}}\Bigg)}{dx}}$

$\sf{\implies \dfrac{d}{dx}\Bigg(\sqrt{x} +\dfrac{1}{\sqrt{x}}\Bigg)}$

$\sf{\implies \dfrac{d}{dx}(\sqrt{x})+\dfrac{d}{dx}\Bigg(\dfrac{1}{\sqrt{x}}\Bigg)}$

$\large{\sf{By\;using\;identity:\;\;\dfrac{d}{dx}(x^{n}) = nx^{n-1}}}$

$\sf{\implies \dfrac{d}{dx}(x)^{\frac{1}{2}}+\dfrac{d}{dx}\Bigg(\dfrac{1}{\sqrt{x}}\Bigg)}$

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

$\large{\boxed{\boxed{\sf{\implies \dfrac{1}{2}x^{-\frac{1}{2}}-\dfrac{1}{2}x^{-\frac{3}{2}}}}}}$

Answer 2

Answer:

$\displatstyle{\dfrac{d}{dx}\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)=\left(-\dfrac{1}{{x}^{5/2}\right)}}$

Explanation:

Given :

$\displaystyle{f(x)=\dfrac{\sqrt{x}+1}{\sqrt{x}}}$

We have to differentiate w.r.t. x

Using quotient rule :

$\displaystyle{\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)=\left(\dfrac{g(x)\times d/dx \ f(x)-f(x)\times d/dx \ g(x)}{\left[g(x)\right]^2}\right)}$

$\displaystyle{\dfrac{d}{dx}\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)=\left(\dfrac{\sqrt x\times d/dx \ \sqrt{x}+1-\sqrt{x}+1\times d/dx \ \sqrt{x}}{\left[\sqrt{x}\right]^2}\right)}\\\\\\\displaystyle{\dfrac{d}{dx}\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)=\left(\dfrac{\sqrt x\times x^{-1/2}+0-\sqrt{x}+1\times x^{-1/2}}{\left[x\right]^2}\right)}$

$\displaystyle{\dfrac{d}{dx}\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)=\left(\dfrac{\dfrac{\sqrt {x}}{\sqrt{x}} - [x^{1/2}\times x^{-1/2}+x^{-1/2}]}{x^2\right)}\\\\\\\displaystyle{\dfrac{d}{dx}\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)=\left(\dfrac{1- 1-x^{-1/2}}{x^2\right)}\\\\\\\displaystyle{\dfrac{d}{dx}\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)=\left(\dfrac{-x^{-1/2}}{x^2\right)}\\\\\\\displaystyle{\dfrac{d}{dx}\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)=\left(-\dfrac{1}{x^2\times{x}^{1/2}\right)}$

$\displaystyle{\dfrac{d}{dx}\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)=\left(-\dfrac{1}{{x}^{5/2}\right)}}$

Thus we get answer .