1. ## find the derivative

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!

2. Hello, finalfantasy!

i'm having some problem differentiating this function.

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

The function is: .$\displaystyle f(x) \;=\;\frac{1}{\sqrt[3]{8x^2 }}$
I can't see how you got $\displaystyle \sqrt{2}$ in there . . .

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

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

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

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

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

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

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