# How do you find the derivative of (1+ln(3x^2))/(1+ln(4x))?

Jan 16, 2018

$\frac{d}{\mathrm{dx}} \left[\frac{1 + \ln \left(3 {x}^{2}\right)}{1 + \ln \left(4 x\right)}\right] = \frac{1 + \ln \left(\frac{16}{3}\right)}{x {\left(1 + \ln \left(4 x\right)\right)}^{2}}$

#### Explanation:

We'll be using the quotient rule for this one:

$\frac{d}{\mathrm{dx}} \left[f \frac{x}{g} \left(x\right)\right] = \frac{f ' \left(x\right) g \left(x\right) - g ' \left(x\right) f \left(x\right)}{{\left(g \left(x\right)\right)}^{2}}$

Let:

$f \left(x\right) = 1 + \ln \left(3 {x}^{2}\right)$

$g \left(x\right) = 1 + \ln \left(4 x\right)$

So:

Use color(blue)(d/dx[ln(u)]=1/u*u'

f'(x)=color(blue)(d/dx[1]+d/dx[ln(3x^2)]=0+1/(3x^2)*d/dx[3x^2]=1/(3x^2)*6x=(6x)/(3x^2)=color(red)(2/x

g'(x)=color(blue)(d/dx[1]+d/dx[ln(4x)]=0+1/(4x)*d/dx[4x]=1/(4x)*4=4/(4x)=color(red)(1/x

By the product rule:

$= \frac{\frac{2}{x} \cdot 1 + \ln \left(4 x\right) - \frac{1}{x} \cdot 1 + \ln \left(3 {x}^{2}\right)}{{\left(1 + \ln \left(4 x\right)\right)}^{2}}$

=((2(1+ln(4x)))/x-((1+ln(3x^2)))/x)/((1+ln(4x))^2

=((2(1+ln(4x))-(1+ln(3x^2)))/x)/((1+ln(4x))^2

Apply the fraction rule for further simplification: $\frac{\frac{a}{b}}{c} = \frac{a}{b} \cdot \frac{1}{c}$

$= \frac{2 \left(1 + \ln \left(4 x\right)\right) - \left(1 + \ln \left(3 {x}^{2}\right)\right)}{x} \cdot \frac{1}{1 + \ln \left(4 x\right)} ^ 2$

$= \frac{2 \left(1 + \ln \left(4 x\right)\right) - \left(1 + \ln \left(3 {x}^{2}\right)\right)}{x {\left(1 + \ln \left(4 x\right)\right)}^{2}}$

We can simplify the numerator if desired

color(blue)(2(1+ln(4x))-(1+ln(3x^2))=2+2ln(4x)-1-ln(3x^2)

Applying several log rules...

color(blue)(2+ln(4x)^2-1-ln(3x^2)

color(blue)(1+ln((4x)^2/(3x^2))=1+ln((16x^2)/(3x^2))=color(red)(1+ln(16/3)

$\therefore \frac{d}{\mathrm{dx}} \left[\frac{1 + \ln \left(3 {x}^{2}\right)}{1 + \ln \left(4 x\right)}\right] = \frac{1 + \ln \left(\frac{16}{3}\right)}{x {\left(1 + \ln \left(4 x\right)\right)}^{2}}$