How do you differentiate $f(x)=ln2^x/(3ln3x)$ ?

Question

1815 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-29T20:24:46

$f'(x)=(3ln2ln3x-3/xln2^x)/(9ln^2(3x)$ .

I will make use of the following rules of differentiation to evaluate this:

The quotient rule : $d/dx[f(x)/g(x)]=(g(x)*f'(x)-f(x)*g'(x))/([g(x)]^2$

Log rule with power rule: $d/dx[lnu(x)]=1/u*(du)/(dx)$

Exponential rule : $d/dxa^x=a^xlna$ .

$therefored/dx((ln2^x)/(3ln3x))=(((3ln3x)*1/2^x*2^xln2)-((ln2^x)*3/(3x)*3))/((3ln3x)^2)$

$=(3ln2ln3x-3/xln2^x)/(9ln^2(3x)$ .

How do you differentiate f(x)=ln2^x/(3ln3x) f(x)=ln2^x/(3ln3x) ?