12-20-2010, 10:20 PM
3rd.
here g(x) = x3 and f(x) =4cos(x)
Y = f(gx) = 4cos(x3)
Let u = x3
Y = 4cos(u)
By chain rule
Dy/dx = dy/du.du/dx = d/du (4cos(u) .d/du (x3)
=(-4sin(u) (3x2)
(-4sin(x3) (3x2)
-12x2sin(x3)
here g(x) = x3 and f(x) =4cos(x)
Y = f(gx) = 4cos(x3)
Let u = x3
Y = 4cos(u)
By chain rule
Dy/dx = dy/du.du/dx = d/du (4cos(u) .d/du (x3)
=(-4sin(u) (3x2)
(-4sin(x3) (3x2)
-12x2sin(x3)
