solving a trig derivative

Question

I'm trying to work this trig derivative but i'm not sure if I'm doing it correctly.

I've edited:

$$ \begin{align} y &= u(a\cos u + b\cot u)\\ \\ y' &=(u)(-a\sin u - b\csc^2 u)+ (a\cos u +b\cot u)(1)\\ &= -au\sin u - bu\csc^2 u + a\cos u + b\cot u\\ \end{align} $$

The derivative of $u$ with respect to $u$ is not $0$. Also, where did $c$ come from in the last line? — John Habert, Feb 20 '14 at 17:19
sorry the c was an error shouldn't be there. i've also updated it — , Feb 20 '14 at 17:26
Unless you are being asked to use trig identities to rewrite into another form, what you have now is correct. — John Habert, Feb 20 '14 at 17:28

amWhy · Accepted Answer · 2014-02-20T17:34:38.623

Seeing that $y$ is a function of $u$, (in other words, you're computing $\frac{dy}{du})$, then $$\frac{d}{du}(u) = 1\neq 0$$

So you cannot drop the right-hand side of the application of the product rule.

I'm also puzzled how/why you introduced the constant $c$ in your move to the final line of your work. Typo?

Re: your edit(s). Your answer is now correct.

But note that you can stop at the the second line $y' = \cdots$, in which you applied the product rule to differentiate $y$. Of course, you'd want to simply drop the factor of $1$ on the right term of your derivative.

After calculating the derivative, you seem to have distributed $u$ over the left term, but then in the next line, you factored out $u$ (which you had just distributed), ending precisely where you were to start with: the expression immediately following $y'$.

oh ok haha. thanks for clearing that up :) – Feb 20 '14 at 17:31 — , Feb 20 '14 at 17:31

solving a trig derivative

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