find the derivative using the difference quotient.?

You are looking for

lim h->0 (f(x+h)-f(x)/h

Looking at the numerator:

(f(x+h)-f(x)= (5/(x+h)^2 + 7) - 5 / (x^2 + 7)) =

(5( (x^2 + 7)) - (x+h)^2 + 7))/((x^2 + 7) (x+h)^2 + 7)) =

5 (x^2 +7 - (x^2 + 2hx + h^2) + 7)/(x^2 + 7) (x+h)^2 + 7) =

5(-2hx -h^2)/(x^2 + 7) (x+h)^2 + 7)

So

lim h->0 (f(x+h)-f(x)/h =

lim h->0 (5(-2hx -h^2)/(x^2 + 7) (x+h)^2 + 7))/h =

lim h->0 (5(-2x - h)/(x^2 + 7) (x+h)^2 + 7)) =

-10x/(x^2 + 7)^2

Quotient rule is used when you have a fraction.

h(x) = f(x) / g(x)

In your case, f(x) = 5 and g(x) = x^2 + 7

and therefore h(x) = 5/(x^2 + 7)

The rules is this

h'(x) = [g(x)f'(x) - g'(x)f(x)] / [g(x)^2]

f'(x) = 0 because f(x) is a constant

g'(x) = 2x

So,

h'(x) = [(x^2 + 7)*0 - (2x*5)] / (x^2 + 7)^2

= -10x / (x^4 + 14x^2 + 49)

MAth is too hard for me...do you ever wonder what weighs more, an ounce of gold, or an ounce of feathers? I can't figure it out.

just pplug it into the quotient rule...you can do it...