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

Answer 1

Elijah Abram

#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)#.

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Answer 2

Scarlett Allison

To differentiate f(x) = ln(2^x) / (3ln(3x)), you can use the quotient rule. The derivative will be:

f'(x) = [((2^x) * ln(2) * 3ln(3x)) - (ln(2^x) * 3) * (1 / (3x))] / (3ln(3x))^2

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.