# How do you find the differential dy of the function y=sqrtx+1/sqrtx?

Sep 4, 2017

Note that $\frac{1}{\sqrt{x}} = {x}^{- \frac{1}{2}}$, and $\sqrt{x} = {x}^{\frac{1}{2}}$, then use the power rule to obtain $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {x}^{- \frac{1}{2}} - \frac{1}{2} {x}^{z \frac{3}{2}}$. Then multiply both sides by dx to get $\mathrm{dy} = \left(\frac{1}{2} {x}^{- \frac{1}{2}} - \frac{1}{2} {x}^{z \frac{3}{2}}\right) \mathrm{dx}$.

#### Explanation:

It will be presumed here that you meant exactly what you said, and want the differential $\mathrm{dy}$ rather than the derivative $\frac{\mathrm{dy}}{\mathrm{dx}}$. If you instead want the derivative, simply ignore the last step.

Recall the power rule states that for any term $a {x}^{b}$, with a and b as constants, the derivative is $\frac{\mathrm{dy}}{\mathrm{dx}} = \left(b\right) a {x}^{b - 1}$. Recall further that $\sqrt{x} = {x}^{\frac{1}{2}}$ and that $\frac{1}{\sqrt{x}} = \frac{1}{x} ^ \left(\frac{1}{2}\right) = {x}^{- \frac{1}{2}}$ (because any term $\frac{p}{x} ^ q = p {x}^{- q}$ by the properties of exponents). Thus, if given the function above, we can find its derivative (and differential) as follows:

$y = \sqrt{x} + \frac{1}{\sqrt{x}} = {x}^{\frac{1}{2}} + {x}^{- \frac{1}{2}}$.

$\to \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} {x}^{\frac{1}{2}} + \frac{d}{\mathrm{dx}} {x}^{- \frac{1}{2}} = \frac{1}{2} {x}^{- \frac{1}{2}} - \frac{1}{2} {x}^{- \frac{3}{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {x}^{- \frac{1}{2}} - \frac{1}{2} {x}^{- \frac{3}{2}}$

This, we now have the derivative. If we want the differential $\mathrm{dy}$...

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {x}^{- \frac{1}{2}} - \frac{1}{2} {x}^{- \frac{3}{2}} \to \mathrm{dy} = \left(\frac{1}{2} {x}^{- \frac{1}{2}} - \frac{1}{2} {x}^{- \frac{3}{2}}\right) \mathrm{dx}$