Question: Find the probability that the birth days of 6 different persons will fall in exactly two calendar months.

Ans.is 341/(12^6)

Here each person has 12 option 
So there are 6 persons .total no. Of ways 12^6

And out of 12 months 2 are randomly selected ..so $12$C$2$ 

B'day of 6 persons fall in 2 months in 2^6 ways.

Therefore requird probability is ($12$C$2$ × 2^$6$)/12^$6$

But not geting appropriate ans.

Answer: 
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
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 $$\left(\frac{\sqrt[3]{4x^3+8}}{(x+2)^5}\right)'=\frac{-16x^3+8x^2-40}{(x+2)^6\sqrt[3]{(4x^3+8)^2}}$$