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I was verifying a larger function derivative on wolfram alpha and came across this derivative:

$\frac{d}{dx} (1-x)^2 = 2(x -1)$

Using the chain rule, I was expecting to get:

$2(1 - x)$

Instead. I trust wolfram alpha can differentiate better than me, so what am I missing?

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3 Answers 3

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You actually didn't use the chain rule correctly: To use the chain rule, you need to also multiply $2(1 - x)$ by $\frac d{dx}(1 - x)= -1$.

In other words, we have $$f(x) = (1 - x)^2 \implies f'(x) = 2(1 - x)\cdot \frac{d}{dx}(1 - x) = -2(1 - x) = 2(x - 1)$$

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  • $\begingroup$ Thanks, I mistakenly thought the derivative of 1-x was 1 instead of -1. $\endgroup$
    – Oscar
    Commented Dec 10, 2013 at 21:59
  • $\begingroup$ Not a problem. I presumed it was a simple lapse of thought involved here. $\endgroup$
    – amWhy
    Commented Dec 10, 2013 at 22:00
  • $\begingroup$ @amWhy: Could use another UV +1 $\endgroup$
    – Amzoti
    Commented Dec 12, 2013 at 4:38
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In more detail: $$ \frac{d\left[(1-x)^2\right]}{dx} = 2 (1-x) \frac{d[1-x]}{dx} = 2 (1-x) (-1) = 2 (x-1). $$

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$\frac{d (1-x)^2}{dx} =2(1-x) \frac{d (1-x)}{dx}=2(1-x)(-1)=2(x-1)$

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