How do you differentiate  y =1/sqrtln(x^2-3x) using the chain rule?

Dec 31, 2015

We can rewrite it as $y = {\left(\ln \left({x}^{2} - 3 x\right)\right)}^{- \frac{1}{2}}$ and use the chain rule, which states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{dx}}$

Explanation:

Renaming:

$y = {u}^{- \frac{1}{2}}$
$u = \ln \left(v\right)$
$v = {x}^{2} - 3 x$

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

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

Substituting $u$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x - 3}{2 {\left(\ln \left(v\right)\right)}^{\frac{3}{2}} \cdot v}$

Substituting $v$:

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

Finally:

(dy)/(dx)=(3-2x)/(2(x^2-3x)sqrt((ln(x^2-3x))^3