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.

It is unlikely that the result of distributing then differentiation will result in something that is accepted. The difference between "not the simplest" and that is an ocean.

But multiple answers are likely correct, because this is a trigonometric expression. There are many equivalent, equally simple ways to represent many such expressions. Makes grading quite difficult. Just because you've never seen or even conceived of an answer does not automatically make it wrong.

I only suggested that as the next step... not the end result.

The next step is far easier if you apply the product rule, rather than going to the basic principles of distributing, differentiating, and factoring.

Applying the product rule leaves the result partially factored. That's why it is used.

Thank you for the correction. I lost track there.

I've, uh, done a bit of calculus in my day.

there is a mistake here. Either your substitution or diff'ing missed something.

the derivative of theta wrt theta is 1. So in these parentheses you should get a -1 and +1...they cancel

my final was y'= -@*cos@/sin^2@ - 2*cos^2@/sin^3@

I can show algebra if you need it.. just lazy right now.

do you have Maple?

It makes you a better person.

I'm sure Buddha or Lao-Tzu would agree.

it isn't really a mistake.. just a poor choice.

I think you have something equivalent... but keep the cot form and you should be able to put your two 'sides' together.

well the form is gf'+fg'

-csc^2@ is the differention of cot@

Have you tried www.wolframalpa.com dont forget to click on show steps

This is the Wiley Plus online math stuff is it? Just curious.

Also, what major are you again? I took a pre-calc/trig course that I should not have...I regret it. A LOT.

y=csc@*(@+cot@)

y'=-@*csc@cot@*(@+cot@)+csc@*[1+(-1-cot^2@)]

=-@*1/sin@*1/tan@ - 2/(sin@*tan^2@)

=-@*cos@/sin^2@ - 2*cos^2@/sin^3@

wolframalpha. com


confusing if you can't do it by hand first.

its great for checking answers, not for doing algebra.

remember the identity you gave in the earlier post?

use that to give each 'side' a cot function. This will allow you to combine them.

I also checked that answer I supplied with software

here is a link to layback's suggestion

it shows some other variations

yeah.. I'm a geek. so what? Here is a wonderful way of doing it by hand using only simple identities and trig.

y = csc(theta)*(theta+cot(theta))

csc(theta)*(theta+cot(theta)) = 1/sin(theta)*(theta+1/tan(theta))

use quotient rule

>
> d(y) = (sin(theta)*0-cos(theta))*(theta+1/tan(theta))/sin(theta)^2+(1+(0-1-tan(theta)^2)/tan(theta)^2)/sin(theta);

> diff(y(x), x) = -theta*cos(theta)/sin(theta)^2-cos(theta)/(sin(theta)^2*tan(theta))-1/(sin(theta)*tan(theta)^2);

> diff(y(x), x) = -theta*cos(theta)/sin(theta)^2-cos(theta)^2/sin(theta)^3-cos(theta)^2/sin(theta)^3;

> diff(y(x), x) = -theta*cos(theta)/sin(theta)^2-2*cos(theta)^2/sin(theta)^3;

Stop it you are going to give me nightmares from Naval Nuclear Power School!!!!!
I have had those memories suppressed for over 20 years..............