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

1 Answer
Trevor Ryan.
Mar 29, 2016

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

Explanation:

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)#.

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