# How do you find the derivative of #y=ln(1-2x)^3#?

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To find the derivative of ( y = \ln(1 - 2x)^3 ), we'll use the chain rule. The derivative of ( \ln(u) ) with respect to ( x ) is ( \frac{1}{u} \cdot \frac{du}{dx} ).

So, applying the chain rule, the derivative of ( y ) with respect to ( x ) is:

[ \frac{d}{dx} \left( \ln(1 - 2x)^3 \right) = \frac{1}{(1 - 2x)^3} \cdot \frac{d}{dx}(1 - 2x)^3 ]

Now, we'll find ( \frac{d}{dx}(1 - 2x)^3 ) using the power rule, which states that the derivative of ( u^n ) with respect to ( x ) is ( n \cdot u^{n-1} \cdot \frac{du}{dx} ). Applying this to ( (1 - 2x)^3 ), we get:

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

Now, finding ( \frac{d}{dx}(1 - 2x) ), which is simply ( -2 ), we substitute it back:

[ \frac{d}{dx}(1 - 2x)^3 = 3 \cdot (1 - 2x)^2 \cdot (-2) = -6(1 - 2x)^2 ]

Putting it all together:

[ \frac{d}{dx} \left( \ln(1 - 2x)^3 \right) = \frac{1}{(1 - 2x)^3} \cdot (-6(1 - 2x)^2) = -\frac{6(1 - 2x)^2}{(1 - 2x)^3} = -\frac{6}{1 - 2x} ]

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