given : g(6)=3 and g prime(6)=-4 B)let p(x)=x^4/g(x),find p prime (6)
This is just the use of the quotient rule in calculus.
If y(x) = u(x)*v(x), then,
y’(x) = (v*u’ – v’*u) / v^2
Taking p(x) = h(x) / g(x), where h(x) = x^4 and h’(x) = 4x^3, then
p’(x) = (g*h’ – g’*h) / g^2
p’(6) = (g(6)*h’(6) – g’(6)*h(6)) / g(6)^2
using g(6) = 3, h’(6) = 4*6^3, g’(6) = -4, h(6) = 6^4, then
p’(6) = (3*4*6^3 – (-4)*6^4) / 3^2
p’(6) = 6^3*(12 + 24) / 9
p’(6) = 216*36/9 = 864
Answer: p’(6) = 864