find the derivative

**finalfantasy** · October 14th 2008, 09:28 AM

i'm having some problem differentiating this function. the answer says (-1/3)x^(-5/3) but the answer i'm getting is (-1/3)(square root 2)(x)^(-5/3)

the function is this:

1/ cube root of 8x^2

thank you!

**Soroban** · October 14th 2008, 09:56 AM

Hello, finalfantasy!

i'm having some problem differentiating this function.

The answer says: $-\frac{1}{3}x^{-\frac{5}{3}}$
but the answer i'm getting is: $-\frac{1}{3}\sqrt{2}\,x^{-\frac{5}{3}}$

The function is: . $f(x) \;=\;\frac{1}{\sqrt[3]{8x^2 }}$

I can't see how you got $\sqrt{2}$ in there . . .

We have: . $f(x) \;=\;\frac{1}{\sqrt[3]{8x^2}} \;=\;\frac{1}{\sqrt[3]{8}\cdot\sqrt[3]{x^2}} \;=\; \frac{1}{2\cdot x^{\frac{2}{3}}} \;=\;\frac{1}{2}x^{-\frac{2}{3}}$

. . $f'(x)\;=\;\left(\frac{1}{2}\right)\left(-\frac{2}{3}\right)x^{-\frac{5}{3}} \;=\;-\frac{1}{3}x^{-\frac{5}{3}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If we didn't simplify first, we can use the Chain Rule . . .

We have: . $f(x) \;=\;\frac{1}{\sqrt[3]{8x^2}} \;=\;\frac{1}{\left(8x^2\right)^{\frac{1}{3}}} \;=\;\left(8x^2\right)^{-\frac{1}{3}}$

Then: . $f'(x) \;=\;-\frac{1}{3}\left(8x^2\right)^{-\frac{4}{3}}(16x) \;=\;-\frac{1}{3}\left(8^{-\frac{4}{3}}\right)\left(x^2\right)^{-\frac{4}{3}}(16x)$

. . $= \;-\frac{1}{3}\left(2^{-4}\right)x^{-\frac{8}{3}}(16x) \;=\;-\frac{1}{3}\cdot\frac{1}{2^4}\cdot x^{\frac{8}{3}}\cdot 16 \cdot x$

. . $= \;-\frac{1}{3}\cdot\frac{1}{16}\cdot 16 \cdot x^{-\frac{8}{3}} \cdot x \;=\;-\frac{1}{3}x^{-\frac{5}{3}}$

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