How do you differentiate y=(x^2-2sqrtx)/x?

Oct 27, 2016

$y ' = \frac{x \sqrt{x} + 1}{x \sqrt{x}}$

Explanation:

The function is differentiated by using the Quotient Rule differentiation

$y = \frac{u \left(x\right)}{v \left(x\right)}$

where $u \left(x\right) = {x}^{2} - 2 \sqrt{x} \mathmr{and} v \left(x\right) = x$

$\textcolor{red}{y ' = \frac{u ' \left(x\right) v \left(x\right) - v ' \left(x\right) u \left(x\right)}{v \left(x\right)} ^ 2}$

Computing $\textcolor{red}{y '}$is determined by computing
$\textcolor{red}{u ' \left(x\right)} \mathmr{and} \textcolor{red}{v ' \left(x\right)}$

$u \left(x\right)$ is a function of polynomials ,it is differentiated using the power rule differentiation rule:
$\textcolor{b l u e}{\left({x}^{n}\right) ' = n {x}^{n - 1}}$

Computing $\textcolor{red}{u ' \left(x\right)}$
Knowing that :sqrtx=x^(1/2)#

$u \left(x\right) = {x}^{2} - 2 \sqrt{x}$
$u ' \left(x\right) = {x}^{2} - 2 {x}^{\frac{1}{2}}$
$u ' \left(x\right) = \textcolor{b l u e}{2 {x}^{1}} - 2 \left(\textcolor{b l u e}{\frac{1}{2} {x}^{- \frac{1}{2}}}\right)$
$u ' \left(x\right) = 2 x - \frac{1}{{x}^{\frac{1}{2}}}$
$\textcolor{red}{u ' \left(x\right) = 2 x - \frac{1}{\sqrt{x}}}$

Computing $\textcolor{red}{v ' \left(x\right)}$
$v \left(x\right) = x$
$\textcolor{red}{v ' \left(x\right) = 1}$

$\textcolor{red}{y ' = \frac{u ' \left(x\right) v \left(x\right) - v ' \left(x\right) u \left(x\right)}{v \left(x\right)} ^ 2}$

$y ' = \frac{\left(2 x - \frac{1}{\sqrt{x}}\right) x - 1 \left({x}^{2} - 2 \sqrt{x}\right)}{x} ^ 2$

$y ' = \frac{2 {x}^{2} - \left(\frac{x}{\sqrt{x}}\right) - {x}^{2} + 2 \sqrt{x}}{x} ^ 2$

$y ' = \frac{{x}^{2} - \left(\frac{x}{\sqrt{x}}\right) + 2 \sqrt{x}}{x} ^ 2$

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

$y ' = \frac{\frac{{x}^{2} \sqrt{x} + x}{\sqrt{x}}}{x} ^ 2$

$y ' = \frac{{x}^{2} \sqrt{x} + x}{{x}^{2} \sqrt{x}}$

$y ' = \frac{x \sqrt{x} + 1}{x \sqrt{x}}$

Oct 27, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 + \frac{1}{x \sqrt{x}}$

Explanation:

As an addition to the other answer, we can also simplify prior to differentiation, and then use the power rule.

$\frac{{x}^{2} - 2 \sqrt{x}}{x} = {x}^{2} / x - 2 \frac{\sqrt{x}}{x}$

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

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

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({x}^{\textcolor{red}{1}} - 2 {x}^{\textcolor{red}{- \frac{1}{2}}}\right)$

$= \textcolor{red}{1} {x}^{1 - 1} - 2 \left(\textcolor{red}{- \frac{1}{2}}\right) {x}^{- \frac{1}{2} - 1}$

$= 1 {x}^{0} - \left(- 1\right) {x}^{- \frac{3}{2}}$

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

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

We could also put it back over a common denominator, giving us the same answer as we would have obtained from the quotient rule.

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

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