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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?)

Jiew Meng
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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

5 Answers5

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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.

Harald Hanche-Olsen
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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 00:27
  • @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 $A+\sqrt[3]{AB^2}-\sqrt[3]{A^2B}-B$ – 2'5 9'2 Feb 25 '12 at 03:17
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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 $Dx^{n} = n.x^{n-1}$

Inquest
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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 00:27
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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 $\frac{dy}{dx}=\frac{1}{3y^2}=\frac{1}{3(\sqrt[3]{x})^2}=\frac{1}{3}x^{-\frac{2}{3}}$.

2'5 9'2
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You can use a similar 'trick' to find the derivative of $ y=\sqrt[n] x $. The limit will involve multiplying the numerator $(x+h)^{1/n} - x^{1/n} $ by an appropriate expression to get $ (x + h) - x $ .

john
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  • I know the answer. I'm asking people if they can find it. But I'll just rephrase it. – john Mar 06 '16 at 23:09
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Note that this works to find the derivative of $x^{1/n}$ where $n$ is a positive integer.

We use $a^n-b^n =(a-b)\sum_{k=0}^{n-1} a^k b^{n-1-k} $.

$\begin{array}\\ \frac{(x+h)^{1/n}-x^{1/n}}{h} &=\frac{(x+h)^{1/n}-x^{1/n}}{h} \frac{\sum_{k=0}^{n-1} ((x+h)^{1/n})^k (x^{1/n})^{n-1-k}}{\sum_{k=0}^{n-1} ((x+h)^{1/n})^k (x^{1/n})^{n-1-k}}\\ &=\frac{(x+h)-x}{h} \frac{1}{\sum_{k=0}^{n-1} ((x+h)^{1/n})^k (x^{1/n})^{n-1-k}}\\ &=\frac{h}{h} \frac{1}{\sum_{k=0}^{n-1} ((x+h)^{1/n})^k (x^{1/n})^{n-1-k}}\\ &=\frac{1}{\sum_{k=0}^{n-1} (x+h)^{k/n} x^{(n-1-k)/n}}\\ \end{array} $

As $h \to 0$,

$\begin{array}\\ \sum_{k=0}^{n-1} (x+h)^{k/n} x^{(n-1-k)/n} &\to \sum_{k=0}^{n-1} x^{k/n} x^{(n-1-k)/n}\\ &= \sum_{k=0}^{n-1} x^{(n-1)/n}\\ &= n x^{(n-1)/n}\\ &= n x^{1-1/n}\\ \end{array} $

Therefore $(x^{1/n})' =\frac1{n x^{1-1/n}} =\frac1{n} x^{1/n-1} $.

marty cohen
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