# What are the first and second derivatives of f(x)=ln(1-x)^(3/2) ?

Dec 30, 2015

Assuming that what you meant was ${\ln}^{\frac{3}{2}} \left(1 - x\right)$, we can apply chain rule twice here.

#### Explanation:

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

Where $f \left(x\right) = {u}^{\frac{3}{2}}$, $u = \ln \left(v\right)$ and $v = 1 - x$.

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

Substituting $u$:

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = - \frac{3 {\left(\ln \left(v\right)\right)}^{\frac{1}{2}}}{2 v}$

Substituting $v$:

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \textcolor{g r e e n}{\frac{3 \sqrt{\ln \left(1 - x\right)}}{2 \left(x - 1\right)}}$

The second derivative falls into the same logic for chain rule in the numerator. However, we have a fraction, which will demand quotient rule.

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

$\frac{\mathrm{df} {\left(x\right)}^{2}}{{d}^{2} x} = \frac{\left(\left(\frac{3}{2} {u}^{- \frac{1}{2}}\right) \left(\frac{1}{v}\right) \left(- 1\right)\right) \left(2 \left(x - 1\right)\right) - 3 \sqrt{\ln \left(1 - x\right)} \left(2\right)}{4 {\left(x - 1\right)}^{2}}$

Substituting $u$:

$\frac{\mathrm{df} {\left(x\right)}^{2}}{{d}^{2} x} = \frac{\left(\left(\frac{3}{2 \sqrt{\ln \left(v\right)}} \left(\frac{1}{v}\right) \left(- 1\right)\right) \left(2 \left(x - 1\right)\right) - 6 \sqrt{\ln \left(1 - x\right)}\right)}{4 {\left(x - 1\right)}^{2}}$

Substituting $v$:

$\frac{\mathrm{df} {\left(x\right)}^{2}}{{d}^{2} x} = \frac{\left(\left(\frac{3}{2 \sqrt{\ln \left(1 - x\right)}} \left(\frac{1}{1 - x}\right) \left(- 1\right)\right) \left(2 \left(x - 1\right)\right) - 6 \sqrt{\ln \left(1 - x\right)}\right)}{4 {\left(x - 1\right)}^{2}}$

$\frac{\mathrm{df} {\left(x\right)}^{2}}{{d}^{2} x} = \frac{\frac{\cancel{-} 6 \cancel{\left(x - 1\right)}}{2 \cancel{\left(1 - x\right)} \sqrt{\ln \left(x - 1\right)}} - 6 \sqrt{\ln \left(1 - x\right)}}{4 {\left(x - 1\right)}^{2}}$

$\frac{\mathrm{df} {\left(x\right)}^{2}}{{d}^{2} x} = \frac{\frac{3 - 6 \ln \left(1 - x\right)}{\sqrt{\ln \left(1 - x\right)}}}{4 {\left(x - 1\right)}^{2}} = \textcolor{g r e e n}{\frac{3 - 6 \ln \left(1 - x\right)}{4 {\left(x - 1\right)}^{2} \sqrt{\ln \left(1 - x\right)}}}$