I am trying to figure out the derivative of $y= \frac{t\sin t}{1+t}$ I know the quotient rules are needed but I think what is confusing is some fairly simple math. Here is what I did.
(1+t)(tcost) - (tsint)(1) this could be wrong but I think it is correct. Anyways what I was confused on was multiplying t into tcost. I forget the rules but is tcost times t should be $t^2\cos$ or something close but I am not sure. Anyways I end up with $ \frac{t\cos t+t^2 \cos t - t \sin t}{(1+t)^2}$ but this is not correct.
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$\begingroup$The product and quotient rules need to be used here. For example, we can first use the product rule $$ \begin{align} \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{t\sin(t)}{1+t}\right) &=\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{t}{1+t}\sin(t)\right)\\ &=\left(\frac{\mathrm{d}}{\mathrm{d}t}\frac{t}{1+t}\right)\sin(t)+\frac{t}{1+t}\left(\frac{\mathrm{d}}{\mathrm{d}t}\sin(t)\right) \end{align} $$ and then use the quotient rule for $\frac{\mathrm{d}}{\mathrm{d}t}\frac{t}{1+t}$. Alternatively, there is nothing wrong with using the quotient rule first and then using the product rule for $\frac{\mathrm{d}}{\mathrm{d}t}(t\sin(t))$.
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