recall the chain rule
fy/fx = dy/du du/dx
#1: u=3x^3
dy/dx = -sin(u) (9x^2) = -9x^2 sin(3x^3)
#2: u = x^4 + e^2x)
dy/dx = 1/u du/dx = (4x^3 + 2e^2x)/ln(x^4+e^2x)
1. Find f '(x) for f(x) = cos (3x3)
2. Find f '(x) for f(x) = ln(x4 + e2x
2 answers
oops #2 is
(4x^3 + 2e^2x)/(x^4+e^2x)
(4x^3 + 2e^2x)/(x^4+e^2x)