How would you go about differentiating 1/sqrt[3]{x} (x^3+tan(x))^4. I said u=1/sqrt[3]{x}(x^3+tan(x)) And y=u^4. Can someone tell me why I'm wrong?
(the sqrt[3]{x} is suppose to signify cubed root of)

Question

How would you go about differentiating 1/sqrt[3]{x} (x^3+tan(x))^4. I said u=1/sqrt[3]{x}(x^3+tan(x)) And y=u^4. Can someone tell me why I'm wrong?
(the sqrt[3]{x} is suppose to signify cubed root of)

homeworksucks · Answer

I have no idea how do this, I'm just going to put it in equation for for you

homeworksucks · Answer

$$\frac{ 1 }{ \sqrt{3} x}(x ^{3}+\tan(x))^{4}$$

homeworksucks · Answer

Is that right?

anonymous · Answer

it's a cubed root and the (x^3+tan(x))^4 is within the cubed root

homeworksucks · Answer

Got it, $$1/\sqrt[3]{x(x^3+\tan(x)^{4}}$$?

anonymous · Answer

Sorry! But it's without the x out in front of the (X^3...)^4

homeworksucks · Answer

Oh, so just the x cubed plus tan(x) to the 4th

homeworksucks · Answer

$$1/\sqrt[3]{(x ^{3}+\tan(x))^{4}}$$

anonymous · Answer

Yes! My mistake. That is correct.

homeworksucks · Answer

Cool! I'll let a person who actually knows a shred of calculus handle it from here

anonymous · Answer

okay, thanks!

zepdrix · Answer

$$\Large \cfrac{1}{\sqrt[3]{(x^3+\tan x)^4}}$$We need derivative? :o
So you were making a substitution to simplify it down or something?

perl · Answer

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