rewrite and differentiate the function without using the quotient rule?

Step by step..........

y = 4/5x^2 = (4x^-2)/5

dy/dx = (-8x^-3)/5 = -8/5x^3

method 1:
y = 4 /(5x^2)

y = 4/5 * 1/x^2

y = 4/5* x^(-2)

dy / dx = y ' = 4/5[(-2)x^(-2-1)]

dy / dx = y ' = 4/5[-2x^(-3]

dy / dx = y ' = (-8/5) (x^-3)

dy / dx = y ' = (-8/5) (1/x^3)
dy / dx = y ' = -8/(5x^3)



method 2:
y = 4 /(5x^2)

lny = ln[4/(5x^2)]

lny = ln4 - ln5 - lnx^2

lny = ln4 - ln5 - 2lnx

d(lny) / dx = y ' / y = 0 - 0 - 2 (1/x )

y ' / y = -2 / x      or     y ' = -2y/x

y ' = -2[4/(5x^2)] / x    (note: y  = 4/(5x^2) )
y ' = -8 / (5x^3)

The other answer is perfectly right, but I thought I would explicate a little  You NEVER need the quotient rule.  Just use the product rule: (uv)'=uv'+u'v.  Treating your problem above with it.  Where u=4 and v=1/(5x^2), (uv)'=uv'+u'v= 4(1/(5x^2))'+0(1/(5x^2))=(4/5)(x^-2)'=(4/5)*(-2)(x^-3)=(-8/5)/(x^3).  If you keep doing calculus, you will probably begin doing that problem in your head without hardly thinking about it and I'd be amazed if anyone on a higher level would do that problem with a quotient rule.  In fact, I can't remember seeing the quotient rule since calc I.  When it comes down to it, the quotient rule really is just the power rule with a little help from the chain rule, just with the necessity of converting a positive denominator to a negatively powered numerator.  Just always remember that power in the denominator is the same as negative power to the same factor as a numerator.  It works especially well for a more complex problem such as (((x-2)^3)/(3x^4))'.  Instead of quotient rule, you just apply the product rule to ((x-2)^3)*(3x^-4).  So as I said, keep the product rule in mind and switch the value of the power of the denominator to a numerator with a negative power and I bet you never have a problem or need the quotient past your class.