# What is the derivative of #f(x)=((x^2-3x)/(2x-3))*ln(x^2-3x)#?

We can approach this by letting

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To find the derivative of the function ( f(x) = \frac{(x^2 - 3x)}{(2x - 3)} \cdot \ln(x^2 - 3x) ), you can use the product rule and the chain rule.

Let ( u = \frac{(x^2 - 3x)}{(2x - 3)} ) and ( v = \ln(x^2 - 3x) ).

Then, using the product rule:

[ f'(x) = u'v + uv' ]

Where: [ u' = \frac{d}{dx}\left(\frac{x^2 - 3x}{2x - 3}\right) ] [ v' = \frac{d}{dx}\ln(x^2 - 3x) ]

Now, calculate ( u' ) and ( v' ) separately.

[ u' = \frac{(2x - 3)(2x) - (x^2 - 3x)(2)}{(2x - 3)^2} ]

[ v' = \frac{1}{x^2 - 3x} \cdot \frac{d}{dx}(x^2 - 3x) = \frac{1}{x^2 - 3x} \cdot (2x - 3) ]

Now, substitute ( u' ) and ( v' ) into the product rule formula and simplify to find ( f'(x) ).

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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.

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.

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.

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