Derivative and solving for '0'

[sdg685]

I'm having a little trouble with this problem. I'm fairly confident that the derivative is correct, but I'm not so certain about the manner in which I solved for '0'. If you could shed some light with some criticism and explanations, that would be greatly appreciated. Thanks! It's problem 5, by the way.

 

[tkhunny]

What is that "((0 cos(x)" stuff in the middle of line 2?

Are you going to get 3pi/2 with that?

 

[sdg685]

What is that "((0 cos(x)" stuff in the middle of line 2?

Are you going to get 3pi/2 with that?

I was attempting to show all my steps of reasoning. That step involves the product, exponent, and chain rules. But, I think that the derivative is correct. The line that has the derivative of f(x) = 0 and and downwards is my attempt at solving for '0'. So, that's pretty much my area of concern. I don't know if those zeroes are correct. Just to clarify, the simplified derivative that I got is: sin(x)cos(x)(3sin(x)+4). From there, I set all of the following as equal to zero: sin(x); cos(x); (3sin(x)+4) . And from those I got the respective values of zero which are shown on the paper. Another problem that I've run into is that if those are all correct, I can't calculate sin(x) = -(4/3) . I'm running into a domain problem. What do I do???? Hahah Anyways, if you could help me out, then you're a **** good person, and to **** with anyone that says otherwise.

 

[Yogi]

It looks pretty solid, in line 2 you don't really need product rule since 1 of the things is a constant, could have just went to power rule directly. Why didn't you do product rule on the first half since you can multiply it by 1? Same reason, it's unnecessary extra steps.

Missing some solutions on cos(x)=0, think unit circle points (cos, sin) so when cos(x)=0 you are at the points (0,1) or (0,-1) which is pi/2, 3pi/2.

Do you need 0 and 0+2npi? You could just say n>=0 and use 0+2npi only. Same for the rest.

What is the range of sin(x)? Knowing this will help you with sin(x) = -(4/3).

 

[sdg685]

It looks pretty solid, in line 2 you don't really need product rule since 1 of the things is a constant, could have just went to power rule directly. Why didn't you do product rule on the first half since you can multiply it by 1? Same reason, it's unnecessary extra steps.

Missing some solutions on cos(x)=0, think unit circle points (cos, sin) so when cos(x)=0 you are at the points (0,1) or (0,-1) which is pi/2, 3pi/2.

Do you need 0 and 0+2npi? You could just say n>=0 and use 0+2npi only. Same for the rest.

What is the range of sin(x)? Knowing this will help you with sin(x) = -(4/3).

The teach didn't specify a range for the problems. He said to solve for all possible Xs. So, I tried my best at doing so by putting in the answers that I did. But, if everything looks good to you, then I'm alright with the answers that I have so far. Thanks so much for the help!

 

[tkhunny]

Oh, that was a product rule treating the constant as a function. Right, don't do that. Way too much trouble.

Your teacher should not have to specify the range fo the sine function. It is something you are expected to know. So, what is it?

 

[sdg685]

Oh, that was a product rule treating the constant as a function. Right, don't do that. Way too much trouble.

Your teacher should not have to specify the range fo the sine function. It is something you are expected to know. So, what is it?

Ah, you're right. I was totally out of it yesterday... But, I guess, I mistook the range for the domain.... (SMH) Anyways, The range is +/- 1. So, that zero doesn't exist. So, that means that the only zeroes at which there is a horizontal slope for that function are where sin(x)=0 and cos(x)=0.