How do you differentiate (x-1)/sqrt(6-x)?

Jan 23, 2018

By using quotient rule:

Explanation:

Let $u = x - 1$ and $v = \sqrt{6 - x}$
$\frac{\mathrm{du}}{\mathrm{dx}} = 1$ and $\frac{\mathrm{dv}}{\mathrm{dx}} = - \frac{1}{2} {\left(6 - x\right)}^{- \frac{1}{2}}$

Plugging this into the formula, you get:

Step one:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(\sqrt{6 - x}\right) \left(1\right) - \left(x - 1\right) \left(- \frac{1}{2}\right) {\left(6 - x\right)}^{- \frac{1}{2}}}{{\left(\sqrt{6 - x}\right)}^{2}}$

Step two: simplify

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(\sqrt{6 - x}\right) + \left(\frac{1}{2}\right) \left(x - 1\right) {\left(6 - x\right)}^{- \frac{1}{2}}}{6 - x}$

Step three: split numerator

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sqrt{6 - x}}{6 - x} + \frac{\left(\frac{1}{2}\right) \left(x - 1\right) {\left(6 - x\right)}^{- \frac{1}{2}}}{6 - x}$

Step Four: combine the $\left(6 - x\right)$ terms using exponent rules

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(6 - x\right)}^{- \frac{1}{2}} + \left(\frac{1}{2}\right) \left(x - 1\right) {\left(6 - x\right)}^{- \frac{3}{2}}$

Step Five: write with positive exponents only

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{\left(6 - x\right)}^{\frac{1}{2}}} + \frac{x - 1}{2 {\left(6 - x\right)}^{\frac{3}{2}}}$

There may be other ways to further simplify the derivative, but this is what I would have done.

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