The following problem may require the use of the following basic trigonometry derivatives. Also, the derivative, d/dx, may be represented by D.
\[\frac{d}{dx}sinx = cosx\]
\[\frac{d}{dx}cosx = sinx\]
\[sin^{2}x + cos^{2}x = 1\]
Differentiate:

\[f(x) = \frac{cosx}{1 + sinx}\]

Solution: Apply the Quotient rule.
\[\frac{df(x)}{dx} = \frac{(1 + sinx) D(cosx)  D(1 + sinx) cosx}{(1 + sinx)^2}\]
\[= \frac{(1 + sinx)(sinx)  (cosx)cosx)}{(1 + sinx)^2}\]
\[= \frac{sinx  sin^2x  cos^2x}{(1 + sinx)^2}\]
\[= \frac{sinx  (sin^2x + cos^2x)}{(1 + sinx)^2}\]
From sin^2x + cos^2x = 1, we get
\[= \frac{sinx  1}{(1 + sinx)^2}\]
\[= \frac{(1 + sinx)}{(1 + sinx)^2}\]
\[= \frac{1}{(1 + sinx)}\]
Final Answer:

\[\frac{1}{(1 + sinx)}\]
