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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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Currently you have $1+\sec \theta$ at the bottom. Is that intended? – André Nicolas Sep 22 '11 at 14:57
I fixed it, that should be the correct problem now. I don't understand what $t\cos\,t\tan\frac{t}{2}$ is – user138246 Sep 22 '11 at 15:00
You almost did it right. The "I think it is correct" part is not. You want $(1+t)$ times the derivative of $t\sin t$ minus $t\sin t$. The derivative of $t\sin t$ is not $t\cos t$, it is $t\cos t +\sin t$ (product rule). – André Nicolas Sep 22 '11 at 15:13
I don't really know what you are trying to say, but is my derivative of tsint wrong? Thinking about it now it should be the derivative of tsint which would be sint(1) + cost(t) is that correct? – user138246 Sep 22 '11 at 15:15
"but is my derivative of tsint wrong" - Yes, exactly. And $\sin t (1) + \cos t (t)$ is correct derivative of $t \sin t$. – Srivatsan Sep 22 '11 at 15:23
up vote 1 down vote accepted

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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