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

Nov 15, 2016

$y ' = - \frac{6}{1 - 2 x}$

#### Explanation:

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$y$ is composite of two functions $\text{ lnx" } \mathmr{and} {\left(1 - 2 x\right)}^{3}$
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Let $u \left(x\right) = \ln x \text{ and " v(x) = (1-2x)^3}$
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Then, $\text{ y=u(v(x))}$
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Differentiating this function is determined by applying chain rule.
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$\textcolor{red}{y ' = u ' \left(v \left(x\right)\right) \times u ' \left(x\right)}$
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color(red)(u'(v(x)) = ?)
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$u ' \left(x\right) = \frac{1}{x}$
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u'(v(x))) = 1/(v(x))
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$\textcolor{red}{u ' \left(v \left(x\right)\right) = \frac{1}{1 - 2 x} ^ 3}$
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color(red)(v'(x) = ?)
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$v ' \left(x\right) = 3 {\left(1 - 2 x\right)}^{2} \left(1 - 2 x\right) '$
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$v ' \left(x\right) = 3 {\left(1 - 2 x\right)}^{2} \left(- 2\right)$
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$\textcolor{red}{v ' \left(x\right) = - 6 {\left(1 - 2 x\right)}^{2}}$
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$\textcolor{red}{y ' = u ' \left(v \left(x\right)\right) \times u ' \left(x\right)}$
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$y ' = \frac{1}{1 - 2 x} ^ 3 \times - 6 {\left(1 - 2 x\right)}^{2}$
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$y ' = \frac{- 6 {\left(1 - 2 x\right)}^{2}}{1 - 2 x} ^ 3$
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$y ' = - \frac{6}{1 - 2 x}$