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What is $f '(x)$ for $f(x)=(x-3)^3$?

I'm thinking it is $3x^2 - 18x + 27$ but my textbook says it is $3x^2 - 18x - 27$

Shaun
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user3184807
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    You are right, the textbook apparently has a typo. – Daniel Fischer Jun 09 '14 at 19:16
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    Textbooks were made by humans. If humans make mistakes, so do textbooks. – Sai82 Jun 09 '14 at 19:25
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    If you want to double check anything you're not sure of, use WolframAlpha. Just type in "derivative of (x-3)^3" into the search box and it will compute it for you. It's much quicker than asking on MSE. – Mr Croutini Jun 09 '14 at 19:27

2 Answers2

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Using the chain rule we have:

$f'(x)=3(x-3)^2\times \frac{d}{dx}(x-3) = 3(x-3)^2$

And $3(x-3)^2 = 3(x^2-6x+9) = 3x^2-18x+27$

You are right. Your textbook is wrong.

Mr Croutini
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Note that $$ f(x)=U^n\\ f^{\prime}(x)=nU^{\prime}U^{n-1} $$ Then $$\begin{align} f^{\prime}(x)&=3(x-3)^{2}\\ &=3(x-3)(x-3)\\ &=3(x(x-3)-3(x-3))\\ &=3(x^2-3x-3x+9)\\ &=3(x^2-6x+9)\\ &=3x^2-18x+27\\ \end{align}$$

Dante
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