Derive (sinx) / (1+tanx)

You use the quotient rule for derivatives since you are dividing.

We will use F (x) for sinx and G (x) for (1+tanx), and H' (x) for the final answer.

So H' (x) = (F' (x) G (x) - F (x) G' (x)) / (G (x)) ^ (2))

The derivative of sinx is cosx so F' (x) = cosx

The derivative of 1+tanx is sec^ (2) x, just like the derivative of tangent (because the derivative of a constant is 0, and by the addition rule for derivatives you would add the derivatives). 0 + sec^ (2) x is sec^ (2) x. So G' (x) is sec^ (2) x.

So with this information just plug F, F', G, and G' into the H' (x) = (F' (x) G (x) - F (x) G' (x)) / (G (x)) ^ (2)) equation and solve.

H' (x) = (cosx) (1+tanx) - (sinx) (sec^ (2) x) / ((1+tanx) ^ (2))