any fans of caclulus?
 

Differentiate the following function.

y = csc(θ) (θ + cot(θ))

I know the derivatives of the trig functions but I am getting stuck when I do the calc. I know its the product rule.. fg'+gf' but after that im stuck

is 'caclulus' a rare, exotic bird?

show us your first step

ok

I will use @ for theta

csc@(1-csc@^2) + (@ + cot@)(-csc@cot@)

then I get confused what to do with the right side.

personally, to simplify this problem, I would gather all the terms by multiplying through.

then change all the uncommon identities (the ones you don't know off the top of your head) to the three common identities (sin cos tan)

then diff.

It's been a while, but why not multiply then solve?

Y = @ csc(@) + cot(@) csc(@)

Now just solve both pieces.

theta/sin(theta)+1/(sin(theta)*tan(theta))

kind of like this.

and there may be more simplification before you diff it all. I think this problem's main point is to simplify as much as possible before diff'ing.

well ive got the differential values of all 6 trig functions.

I multipled them through but im just confused what to do with the right side

with the -csc^2@-csc@cot^2@ result

-csc^2@ is equivelent to -1-cot^2@ [ csc^2@ - cot^2@ = 1]

I just multiplied out, like craig said, and differentiated separately. I got

1csc(theta)-(theta)csc(theta)cot(theta)-csc(theta)cot^2(theta)-csc^3(theta)

that works.. its still a little unclear. You will have to do some algebra to simplify then use identities to simplify again.

I am not certain how far you are expected to go... but Proffs usually want to see the work and simplification and present it to them in the simplest form.

Simplifying as much as possible before diff'ing usually improves the outcome of the differentiation.

any monkey can plug the problem into a computer and write the complicated, inelegant answer.

here is the sucky part.. the HW is done online so showing work doesnt matter to the computer. I've got to get it to simplest form lol

Because then, you have to factor it again to make it useful.

Use the product rule, but write it explicitly. Perhaps using the ' character to indicate the derivative for each part. Until you are comfortable, don't try to just emit the entire answer. That's too "formal" (i.e., using a formula to do mathematics) to be useful for learning.

There still may be factoring to do. Likely in fact when trigonometric functions are involved, but it's more than twice the work to distribute the multiplication over the addition before taking the derivative.

As I said, now factor that beast.

That expression is entirely useless in its current form.

I love this guy...

How many tries do you get online? Usually any equivalent answer will work, even if not simplest. At least that is how I remember it. I like online homework since I didn't have to worry about the whole "showing your work" business or grading errors.