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I need to finding derivative of $\sqrt[3]{x}$ using only limits

So following tip from yahoo answers: I multiplied top and bottom by conjugate of numerator

$$\lim_{h \to 0} \frac{\sqrt[3]{(x+h)} - \sqrt[3]{x}}{h} \cdot \frac{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}$$

$$= \lim_{h \to 0} \frac{x+h-x}{h(\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2})}$$

$$= \lim_{h \to 0} \frac{1}{\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}}$$

$$= \frac{1}{\sqrt[3]{x^2} + \sqrt[3]{x^2}}$$

$$= \frac{1}{2 \sqrt[3]{x^2}}$$

But I think it should be $\frac{1}{3 \sqrt[3]{x^2}}$ (3 instead of 2 in denominator?)

UPDATE

I found that I am using the wrong conjugate in step 1. But this (wrong) conjugate gives the same result when I multiply the numerator by it. So whats wrong with it? (I know its wrong, but why?)

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You made a mistake in the first step. The numerator in the second line should be $(x+h)^{2/3}-x^{2/3}$. –  Harald Hanche-Olsen Feb 24 '12 at 13:55
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The first step is wrong. Yahoo isn't the best place to take hints from. Also, the guy on yahoo is tending h to $\infty$ instead of $0$. –  Inquest Feb 24 '12 at 13:58
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@Harald: Yes, there is the computational mistake that you mentioned, but I would emphasize (as you hint at in your answer) that the problem is motivational: there is a misconception (thanks, yahoo) that is motivating the OP to multiply by the wrong thing (on top and bottom). –  The Chaz 2.0 Feb 24 '12 at 15:26
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3 Answers 3

up vote 5 down vote accepted

Here is a hint: Use the identity $(a^3-b^3)=(a-b)\cdot(a^2+ab+b^2)$ with $a$, $b$ being suitable cube roots. Otherwise, the method is similar to the one you tried.

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That identity was in my notes, but I wonder why the conjugate $\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}$ does not work? When I multiply out the numerator, it gives the same result. –  Jiew Meng Feb 25 '12 at 0:27
    
@Jiew Then you are not multiplying out correctly. $(\sqrt[3]{A}-\sqrt[3]{B})(\sqrt[3]{A^2}+\sqrt[3]{B^2})$ does not multiply out to $A-B$, but rather to $A+\sqrt[3]{AB^2}-\sqrt[3]{A^2B}-B$ –  alex.jordan Feb 25 '12 at 3:17
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$$\lim_{h \to 0} \frac{{(x+h)^{\frac{1}{3}}} - {x}^{\frac{1}{3}}}{h} $$ $$=\lim_{h \to 0} \frac{{(x+h)^{\frac{1}{3}}} - {x}^{\frac{1}{3}}}{h} \cdot \frac{(x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3}}{(x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3}} $$

$$=\lim_{h \to 0} \frac{x+h-x}{h((x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3})}$$ $$=\lim_{h \to 0} \frac{1}{(x)^{2/3} + x^{1/3}(x+h)^{1/3} + (x+h)^{2/3}}$$ $$=\frac{1}{(x)^{2/3} + x^{1/3}(x)^{1/3} + (x)^{2/3}}$$ $$=\frac{1}{3x^{2/3}}$$ $$=\frac{x^{-2/3}}{3}$$ As obtained from the $Dx^{n} = n.x^{n-1}$

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As I said. Some of your $(x+h)^{2/3}$ should have been $x^{2/3}$. But really, I find it misguided to give complete answers to questions with the homework tag. –  Harald Hanche-Olsen Feb 24 '12 at 14:17
    
@HaraldHanche-Olsen, I'm so sorry! I didn't notice the homework tag. I edited the answer but. –  Inquest Feb 24 '12 at 14:23
    
Don't beat yourself up over it. Mistakes happen. So long as it wasn't on purpose, it's no big deal. –  Harald Hanche-Olsen Feb 24 '12 at 14:53
    
That identity was in my notes, but I wonder why the conjugate $\sqrt[3]{(x+h)^2} + \sqrt[3]{x^2}$ does not work? When I multiply out the numerator, it gives the same result. –  Jiew Meng Feb 25 '12 at 0:27
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I understand that the point of this exercise is to apply the limit definition of the derivative to a function where the limit calculation is "tricky". But it's worth noting that if $F(x,y)=0$ identically (as in $y-\sqrt[3]{x}=0$ in this problem) then $\frac{dy}{dx}=\frac{1}{\frac{dx}{dy}}$.

So given that $x=y^3$, we have that $\frac{dx}{dy}=3y^2$ (either using the power rule or a simpler limit computation). That makes $\frac{dy}{dx}=\frac{1}{3y^2}=\frac{1}{3(\sqrt[3]{x})^2}=\frac{1}{3}x^{-\frac{2}{3}}$.

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