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

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Answer 1 · 2018-01-16T20:51:17

$d/dx[(1+ln(3x^2))/(1+ln(4x))]=(1+ln(16/3))/(x(1+ln(4x))^2)$

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

$d/dx[f(x)/g(x)]=(f'(x)g(x)-g'(x)f(x))/((g(x))^2)$

Let:

$f(x)=1+ln(3x^2)$

$g(x)=1+ln(4x)$

So:

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

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

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

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

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

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

Apply the fraction rule for further simplification: $(a/b)/c=a/b*1/c$

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

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

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...

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

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

$:.d/dx[(1+ln(3x^2))/(1+ln(4x))]=(1+ln(16/3))/(x(1+ln(4x))^2)$

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