if f(x)=3/x+x^3 and g(x)=1-x^2 find d/dx f(g(x)).
df/dg = d/dg(3/g + g^3) = -3/g^2 + 3g^2
d/dx f(g(x)) = df/dg * dg/dx = -2x(-3/(1-x^2)^2 + 3(1-x^2)^2)
= 6x/(1-x^2)^2 (1 - (1-x^2)^4)
check:
f(g) = 3/g + g^3
f(g(x)) = 3/(1-x^2) + (1-x^2)^3
d/dx (f(g)) = 6x/(1-x^2)^2 - 6x(1-x^2)^2
= 6x/(1-x^2)^2 (1 - (1-x^2)^4)