If [tex]f(x)=4-x^2[/tex], then
[tex]f(x+h)=4-(x+h)^2=4-(x^2+2xh+h^2)=4-x^2-2xh-h^2[/tex]
So
[tex]\dfrac{f(x+h)-f(x)}h=\dfrac{(4-x^2-2xh-h^2)-(4-x^2)}h=-\dfrac{2xh+h^2}h[/tex]
Then for [tex]h\neq0[/tex], we can cancel common terms:
[tex]\dfrac{f(x+h)-f(x)}h=-2x+h[/tex]
