What is the derivative of sqrt((7x+2) / (6x−3)) ?

May 2, 2017

$= - \setminus \frac{11 \setminus \sqrt{6 x - 3}}{6 {\left(3 x - 1\right)}^{2} \setminus \sqrt{7 x + 2}}$

Explanation:

We can rewrite the function as so
${\left(\setminus \frac{7 x + 2}{6 x - 3}\right)}^{\frac{1}{2}}$

We can then use the chain rule
1/2(\frac{7x+2}{6x-3})^{-1/2}=\frac{1}{2(\frac{7x+2}{6x-3})^{1/2}

We then have to multiply this by the function inside the radical. To do this, we have to use the quotient rule (low dee high minus high dee low down below the square must go).

$\setminus \frac{\left(6 x - 3\right) \left(7\right) - \left(7 x + 2\right) \left(6\right)}{{\left(6 x - 3\right)}^{2}}$
$= \setminus \frac{42 x - 21 - 42 x - 12}{{\left(6 x - 3\right)}^{2}}$
$= - \setminus \frac{33}{{\left(6 x - 3\right)}^{2}}$
$= - \setminus \frac{33}{9 {\left(2 x - 1\right\}}^{2}}$
=-\frac{11}{3(2x-1)^2

Multiplying the two derivatives we got, we get
$\setminus \frac{1}{2 {\left(\setminus \frac{7 x + 2}{6 x - 3}\right)}^{\frac{1}{2}}} \left(- \setminus \frac{11}{3 {\left(2 x - 1\right)}^{2}}\right)$
$= - \setminus \frac{{\left(6 x - 3\right)}^{\frac{1}{2}}}{2 {\left(7 x + 2\right)}^{\frac{1}{2}}} \setminus \cdot \setminus \frac{11}{3 {\left(2 x - 1\right)}^{2}}$
$= - \setminus \frac{11 {\left(6 x - 3\right)}^{\frac{1}{2}}}{6 {\left(3 x - 1\right)}^{2} {\left(7 x + 2\right)}^{\frac{1}{2}}}$