# How do you find derivative of #y=(-2lnx)/(3x-1)#?

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To find the derivative of (y = \frac{-2\ln(x)}{3x-1}), you can use the quotient rule for derivatives, which states that the derivative of a quotient (y = \frac{u}{v}) is given by (y' = \frac{u'v - uv'}{v^2}), where (u) and (v) are functions of (x), and (u') and (v') are their respective derivatives.

Here, (u = -2\ln(x)) and (v = 3x-1).

**Differentiate (u = -2\ln(x)):**

[u' = \frac{d}{dx}(-2\ln(x)) = -2 \cdot \frac{1}{x} = -\frac{2}{x}]

**Differentiate (v = 3x - 1):**

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

**Apply the quotient rule:**

[y' = \frac{u'v - uv'}{v^2} = \frac{\left(-\frac{2}{x}\right)(3x-1) - (-2\ln(x))(3)}{(3x-1)^2}]

[y' = \frac{-\frac{6x-2}{x} + 6\ln(x)}{(3x-1)^2}]

[y' = \frac{-6 + \frac{2}{x} + 6\ln(x)}{(3x-1)^2}]

Thus, the derivative of (y = \frac{-2\ln(x)}{3x-1}) with respect to (x) is (y' = \frac{-6 + \frac{2}{x} + 6\ln(x)}{(3x-1)^2}).

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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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