Would this be a correct differentiation :
$d(\sqrt[3]{1-x^3})=(-x^2*dx)/(1-x^3)^\frac{2}{3}$
I don't know why, but the textbook shows me this answer :
$-\sqrt[2]{(1-x^3)^2}*dx$
Can somebody guide me ? Thank yoU!
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I don't know why either: it looks like they forgot to print the "chain-rule" factor. There should be a factor of $ \ x^2 \ $ in the answer. – colormegone May 04 '14 at 01:38
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So, it would be an error ? ... – May 04 '14 at 01:39
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1Sure would -- somehow the minus sign made it in, so I think there's an omission "typo". – colormegone May 04 '14 at 01:41
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Yeah, I think that's it. Thank you for your time ! – May 04 '14 at 01:42
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Oh, by the way, I just noticed that it's not a cuberoot that's given but a square root. what.... (In the textbook) – May 04 '14 at 01:44
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Where is that? You have a cube-root typed in your post. – colormegone May 04 '14 at 01:46
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Yeah, that was an error from my part. What the textbook says is the following : -squareroot((1-x^3)^2)*dx – May 04 '14 at 01:47
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Well, it looks like a mangled result: a square-root there is a bit silly. You have carried out the differentiation correctly. – colormegone May 04 '14 at 01:49
2 Answers
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$$f(x)=\sqrt[3]{x}=x^{\frac{1}{3}}$$ $$g(x)=1-x^3$$
Clearly $(f \circ g)(x)=f(g(x))=\sqrt[3]{1-x^3}$
Then by the chain rule: $$[f(g(x))]'=f'(g(x))g'(x)$$ Replacing: $$[f(g(x))]'=\frac{1}{3}\frac{1}{(1-x^3)^{\frac{2}{3}}}(-3x^2)=\frac{-x^2}{(1-x^3)^{\frac{2}{3}}}$$
So you're right
rlartiga
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Let $y=(1-x^3)^{1/3}$
$$\dfrac {dy}{dx} = 1/3(1-x^3)^{-2/3}(-3x^2)$$
$$\dfrac {dy}{dx} = -x^2/(1-x^3)^{2/3}$$
so yes you are correct!
The book probably had a typing mistake
