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I need to differentiate this:

$$v(x) = \frac{\sqrt[3]{x-1}}{(x+2)^2}$$

I used this formula:

$$ \frac{f'(x)g(x) - f(x)g'(x)}{g^2(x)}$$

Where:

$$ f'(x) = \frac{1}{3\sqrt[3]{(x-1)^2}}$$ and $$ g'(x) = 2(x+2)$$

so my result is:

$$ \frac{ \frac{1}{3\sqrt[3]{(x-1)^2}}(x+2)^2 - \sqrt[3]{(x-1)} \cdot2(x+2)}{(x+2)^4} = \frac{ \frac{1}{3\sqrt[3]{(x-1)^2}}(x+2)^2 - 2\sqrt[3]{(x-1)}(x+2) }{(x+2)^4} = \frac{ (x+2)^2 - 2\sqrt[3]{(x-1)}(x+2) }{3\sqrt[3]{(x-1)^2}(x+2)^4} = \frac{ (x+2) \big[ (x+2) - 2\sqrt[3]{(x-1)} \big] }{3\sqrt[3]{(x-1)^2}(x+2)^4} = \frac{ (x+2) - 2\sqrt[3]{(x-1)} }{3\sqrt[3]{(x-1)^2}(x+2)^3}$$

but according to wolframalpha the result should be:

$$\frac{ 8 - 5x }{3\sqrt[3]{(x-1)^2}(x+2)^3}$$

mathlove
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Christian Giupponi
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    "solve this"... no: differentiate this. Your notation is poor because you use $f(x)$ for two different things... use $u(x)$ and $v(x)$. – JP McCarthy Jun 25 '15 at 09:59

1 Answers1

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Multiplying top and bottom of $$\frac{\frac{1}{3\sqrt[3]{(x-1)^2}}(x+2)^2-2\sqrt[3]{x-1}\ (x+2)}{(x+2)^4}$$ by $3\sqrt[3]{(x-1)^2}$ gives you $$\frac{(x+2)^2-2\sqrt[3]{x-1}\ (x+2)\cdot\color{red}{3\sqrt[3]{(x-1)^2}}}{3\sqrt[3]{(x-1)^2}(x+2)^4}.$$

mathlove
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