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What is the derivative of #f(x)=ln(2x^2-4x)/x#?

Answer 1

Cameron Alexander

#f'(x) = ((2x^2)/(x^2 - 2) - ln(2x^2 - 4x))/(x^2)#

Start by using the chain rule to differentiate #y = ln(2x^2 - 4x)#.

Let #y = lnu# and #u = 2x^2 - 4#.

#dy/dx = 1/u xx 4x = (4x)/(2x^2 - 4) = (2(2x))/(2(x^2 - 2)) = (2x)/(x^2 - 2)#

By the quotient rule:

#f'(x) = ((2x)/(x^2 - 2) xx x - 1 xx ln(2x^2 - 4x))/(x)^2#

#f'(x) = ((2x^2)/(x^2 - 2) - ln(2x^2 - 4x))/(x^2)#

Hopefully this helps!

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Answer 2

Ezra Adams

The derivative of (f(x) = \frac{\ln(2x^2 - 4x)}{x}) is (f'(x) = \frac{2 - 4x}{x(2x^2 - 4x)} - \frac{\ln(2x^2 - 4x)}{x^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.