1. ## Chain Rule W/Trig

Find the derivative of $y = \left(-1 - \frac{csc \Theta}{2} - \frac{\Theta^2}{4}\right)^2$

So I'm guessing here you just use the chain rule to find the derivative. I get:
$y' = 2 \left(-1 - \frac{1}{2} csc \Theta - \frac{1}{4} \Theta^2\right)\left(\frac{1}{2}csc \Theta cot\Theta - \frac{1}{2}\Theta\right)$

If I know I did this much correct so far I can go on and simplify it.

2. Originally Posted by VitaX
Find the derivative of $y = \left(-1 - \frac{csc \Theta}{2} - \frac{\Theta^2}{4}\right)^2$

So I'm guessing here you just use the chain rule to find the derivative. I get:
$y' = 2 \left(-1 - \frac{1}{2} csc \Theta - \frac{1}{4} \Theta^2\right)\left(\frac{1}{2}csc \Theta cot\Theta - \frac{1}{2}\Theta\right)$

If I know I did this much correct so far I can go on and simplify it.
Looks okay to me.

3. Originally Posted by VitaX
Find the derivative of $y = \left(-1 - \frac{csc \Theta}{2} - \frac{\Theta^2}{4}\right)^2$

So I'm guessing here you just use the chain rule to find the derivative. I get:
$y' = 2 \left(-1 - \frac{1}{2} csc \Theta - \frac{1}{4} \Theta^2\right)\left(\frac{1}{2}csc \Theta cot\Theta - \frac{1}{2}\Theta\right)$

If I know I did this much correct so far I can go on and simplify it.

Well, csc x = 1/sinx, so (csc x)' = -cos x/sin^2(x). Do you really think this is nastier than -csc x*cot x??? Of course, you save yourself a quotient...

Anyway, the derivation looks fine to me.

Tonio

4. Originally Posted by tonio
Well, csc x = 1/sinx, so (csc x)' = -cos x/sin^2(x). Do you really think this is nastier than -csc x*cot x??? Of course, you save yourself a quotient...

Anyway, the derivation looks fine to me.

Tonio
The derivative of $\csc x$ is quoted in all the reference works and textbooks that I've ever come across as $-\csc x \cot x$. IMO it's perfectly appropriate, and I actually find it easier to learn like that. As you say, it saves a quotient.

To take it to its logical conclusion, there's no point in learning it anyway, because you can always use the quotient rule on $\cos x / \sin x$.