# What are the first and second derivatives of f(x)=2ln(sqrt((4x-4)/(6x+10))) ?

Dec 25, 2015

We'll need the chain rule and the quotient rule here.

#### Explanation:

• Chain rule states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{dx}}$

• Quotient rule states that for $y = f \frac{x}{g} \left(x\right)$, then $y ' = \frac{f ' g - f g '}{g} ^ 2$

Renaming $u = \sqrt{v}$ and $v = \frac{4 x - 4}{6 x + 10}$, we have the new $f \left(x\right) = 2 \ln \left(u\right)$ and can now start, but let's do it step-by-step.

$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{2}{u}$

$\frac{\mathrm{du}}{\mathrm{dv}} = \frac{1}{2 \sqrt{v}}$

$\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{\left(4\right) \left(6 x + 10\right) - \left(4 x - 4\right) \left(6\right)}{6 x + 10} ^ 2 = \frac{24 x + 40 - 24 x + 24}{6 x + 10} ^ 2 = \frac{64}{6 x + 10} ^ 2$

Aggregating:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{128}{u \cdot 2 \sqrt{v} \cdot {\left(6 x + 10\right)}^{2}}$

Substituting $u$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{128}{\sqrt{v} 2 \sqrt{v} {\left(6 x + 10\right)}^{2}} = \frac{128}{2 v \cdot {\left(6 x + 10\right)}^{2}}$

Substituting $v$ (Holy Jesus!):

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{128}{2 \left(\frac{4 x - 4}{\cancel{6 x + 10}}\right) {\left(6 x + 10\right)}^{\cancel{2}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{64}{\left(4 x - 4\right) \left(6 x + 10\right)} = \frac{64}{24 {x}^{2} + 16 x - 40}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{8}{3 {x}^{2} + 2 x - 5}$

Now, the second derivative:

$\frac{{\mathrm{dy}}^{2}}{{d}^{2} x} = \frac{\left(0\right) \left(3 {x}^{2} + 2 x + 5\right) - \left(8\right) \left(6 x + 2\right)}{3 {x}^{2} + 2 x - 5} ^ 2 = - \frac{48 x + 16}{3 {x}^{2} + 2 x - 5} ^ 2$