Find the derivative of $y= \frac{(4x^3 +8)^{\frac{1}{3}}}{(x+2)^5}$ How can we find the derivative of 
$y= \frac{(4x^3 +8)^{\frac{1}{3}}}{(x+2)^5}$?
so far this is what I have done, and am confused about what to do after?

sorry for the messy handwriting
 A: I'm assuming you mean
How do we find the derivative of $y= \frac{(4x^3 +8)^{\frac{1}{3}}}{(x+2)^5}$?
The quotation rule:
$\frac{f(x)}{g(x)}=\frac{f'(x)g(x)-g'(x)f(x)}{g^2(x)}$
The chain rule:
$f(g(x))=f'(g(x))g'(x)$
So to solve the question, we apply the quotation rule first
Letting
$$(4x^3 +8)^{\frac{1}{3}}=f(x)$$
$$(x+2)^5=g(x)$$
By applying the quotation rule, we get
$y'=\frac{f'(x)g(x)-g'(x)f(x)}{g^2(x)}$
You can use the chain rule to differentiate both $f(x)$ and $g(x)$
$f'(x)=\frac{1}{3}(4x^3+8)^{-\frac{2}{3}}12x^2$
$g'(x)=5(x+2)^4$
You can plug all of these into $y'=\frac{f'(x)g(x)-g'(x)f(x)}{g^2(x)}$ to get the derivative
A: Use this relation $$( \frac{f}{g})'=\frac{(f')g-(g')f}{g^2}$$
with
$$f(x)=(4x^3+8)^{1/3}$$
and
$$g(x)=(x+2)^5. $$
Also use the chain rule.
A: You have optionnally two equivalent ways: $$(\frac{u}{v})'=(\frac{u'v-uv'}{v^2})\\(\frac 1v\cdot u)'=(\frac 1v)'\cdot u+(\frac 1v)\cdot u'$$
You must get after some care in calculations $$\left(\frac{\sqrt[3]{4x^3+8}}{(x+2)^5}\right)'=\frac{-16x^3+8x^2-40}{(x+2)^6\sqrt[3]{(4x^3+8)^2}}$$
A: Hint
When you face products, quotients, powers, .. , you can make life much easier using logarithmic differentiation. In your case $$y = \frac{\sqrt[3]{4x^3+8}}{(x+2)^5}\implies \log(y)=\frac 13 \log(4x^3+8)-5\log(x+2)$$ Now, differentiate $$\frac{y'}y=\frac 13 \times\frac {12x^2}{4x^2+8}-5\times\frac 1{x+2}= ??$$
Simplify (reducing to same denominator) and use $y'=y\times \frac{y'}y$ and simplify the powers.
