PLEASE HELP CALCULUS:

Use the Generalized Power Rule to find the derivative of the function.

y=[1/(w^3-1)]^8

Question

PLEASE HELP CALCULUS:

Use the Generalized Power Rule to find the derivative of the function. 

y=[1/(w^3-1)]^8

anonymous · Answer

$$y=\left( \frac{ 1 }{ w^3-1 } \right) ^8$$

nincompoop · Answer

okay you need quotient and chain rule

nincompoop · Answer

rules*

anonymous · Answer

Or you can make it to the power of -1 and use chain rule/ power rule st00f, as it suggests.

nincompoop · Answer

$$
y = \frac{1^8}{(w^3-1)^8}
$$

nincompoop · Answer

perform quotient rule and power rule

anonymous · Answer

so this was my answer
$$y=8(w ^{-3}-1)^7(-3w ^{-4})$$

nincompoop · Answer

I don't really care about the answer that you said was already wrong

anonymous · Answer

Your derivative is wrong, just do what nin asks D: he's fat

anonymous · Answer

ok, I dont understand what u want me to with that?

nincompoop · Answer

$$
u = 1^8 \
u' = 0 \
v = (w^3-1)^8 \
v' = 8(w^3-1)^7 \times 2w^2 \


y' = \frac{u'v - uv'}{v^2}
$$

nincompoop · Answer

sorry

anonymous · Answer

oh the quotient rule. I forgot about that.

nincompoop · Answer

$$
v' = 8(w^3-1)^7 \times 3w^2
$$

nincompoop · Answer

went trigger happy with the 2 there 
sorry for the typo

anonymous · Answer

ok the reason I didn't approach with the quotient rule was because with the demonstration of webassign, it was showing me to approach the equation by y=n(blah blah)^n-1(dy/dx)