Answer
$y'=\dfrac{\cos t+\sin t-\sin t\sec^{2}t}{(1+\tan t)^{2}}$
Work Step by Step
$y=\dfrac{\sin t}{1+\tan t}$
Differentiate using the quotient rule:
$y'=\dfrac{(1+\tan t)(\sin t)'-(\sin t)(1+\tan t)'}{(1+\tan t)^{2}}=...$
$...=\dfrac{(1+\tan t)(\cos t)-(\sin t)(\sec^{2}t)}{(1+\tan t)^{2}}=...$
Evaluate the products in the numerator:
$...=\dfrac{\cos t+\cos t\tan t-\sin t\sec^{2}t}{(1+\tan t)^{2}}$
We know $\tan t=\dfrac{\sin t}{\cos t}$, so we substitute and simplify:
$...=\dfrac{\cos t+(\cos t)(\dfrac{\sin t}{\cos t})-\sin t\sec^{2}t}{(1+\tan t)^{2}}=...$
$...=\dfrac{\cos t+\sin t-\sin t\sec^{2}t}{(1+\tan t)^{2}}$
