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To solve this the book does this:

$\frac{d}{dx}(3x)^{-4}$ $=$ $\frac{d}{dx}\frac{1}{(3x)^{4}} = \frac{d}{dx}\frac{1}{81x^{4}} = \frac{-4}{81}\cdot x^{-5}$

But when I tried doing this:

$\frac{d}{dx}(3x)^{-4} = -4(3x)^{-5} = \frac{-4}{243}\cdot x^{-5}$

How come my method doesn't work? Am I missing something or is this a mistake?

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    $\begingroup$ Hint: chain rule $\endgroup$ Oct 18 '15 at 1:22
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You have to use the chain rule in your method. So you have to multiply by the derivative of $3x$.

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Here's another way to think about it... We have a constant coefficient $3^{-4}$. Let's bring it in front of the derivative before applying the power rule.

$$\begin{align*} \frac{d}{dx}(3x)^{-4} &= (3)^{-4}\frac{d}{dx}x^{-4} \\ &= (3)^{-4}(-4)x^{-5} \\ &= \frac{-4}{81}x^{-5} \end{align*}$$

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