Question

Find dy/dx of y=cos(6x+5)

Answer 1

I know its the chain rule, i already got. Cos(6x+5) * 6

Answer 2

That... should be it.

Answer 3

Using u substitution, but i'm not sure if i did that quite right, Does cos become -sin?

Answer 4

Oh yes

Answer 5

Wait.

Answer 6

So -sin(6x+5)*6? Can that simplify more?

Answer 7

\[f'(g(x))g'(x) = -\sin(6x + 5)\cdot 6\]Yes, you may use \(-\sin(x) = \sin(-x)\) but you don't need it. So leave it at \(-6\sin(6x + 5)\)

Answer 8

Think you got your trig messed up there lol.

Answer 9

Derivative of Sin is cos, and cos is -sin.

Answer 10

I guess that's exactly what I did there...?

Answer 11

Well you said -sin=sin :P

Answer 12

Also the 6 moves up from with the -sin?

Answer 13

No... I said \(-\sin(x) = \sin(-x)\). And yeah, \(-\sin(6x + 5) \cdot 6 = -6\sin(6x+5)\)

Answer 14

Oh i see, You're right thats a trig identity, but not for derivatives.

Answer 15

Thanks for the help again : ).

Answer 16

:-\ I think \(-\sin(x) = \sin(-x)\) everywhere. You don't need to use that here though. Hmm

Answer 17

Yeah it probably is, just didnt need it for this.

Answer 18

You still here, got another question?

Answer 19

Yes.

Answer 20

I can't tell if i need the chain rule, or the sum rule here. or both?

Answer 21

y=(cscx+cotx)^-1

Answer 22

So in a way you want to do\[\dfrac{1}{\csc(x) + \cot(x)}\]Right?

Answer 23

I guess technically thats the same thing yeah.

Answer 24

Seems to make it more complicated imo.

Answer 25

Or do i use the product rule now?

Answer 26

Quotient i mean*

Answer 27

I don't know the derivatives of \(\cot(x)\) or \(\csc(x)\) so I'd begin with simplification.\[\dfrac{1}{\frac{\cos(x)}{\sin(x)}+\frac{1}{\sin(x)}} = \dfrac{1}{\frac{1 + \cos(x)}{\sin(x)}} = \dfrac{\sin(x)}{1 + \cos(x)}\]

Answer 28

Would that make it simple?

Answer 29

Well the derv of csc(x)=-cscx*cotx and cot(x)=-cscx^2

Answer 30

I don't memorize those :-|

Answer 31

Those are just basic trig functions. I don't think we can do it like that, might lead to the same answer, but ultimately we need dy/dx.

Answer 32

I haven't memorized them yet either, need to by tomorrow though :P

Answer 33

Yeah, you may do it that way. Never stopped you :-P

Answer 34

Those are just basic trig functions. I don't think we can do it like that, might lead to the same answer, but ultimately we need dy/dx.

Answer 35

Well then back to my original question. Do i use the chain rule or how would i go about doing it?

Answer 36

You can use the quotient rule.

Answer 37

Too many options i feel like i'm doing something wrong hmmm

Answer 38

How do you want to do it?

Answer 39

Not sure lol. The correct way?

Answer 40

All ways are correct :-P

Do you want to continue from \(\dfrac{\sin(x)}{1 + \cos(x)}\)?

Answer 41

Sure we can. I was doing the quotient rule on my paper.

Answer 42

Actually wait, is this right so far.

Answer 43

\[y=\frac{ \csc(x)\cot(x)+\csc^2x }{ (\csc(x)\cot(x))^2 }\]

Answer 44

Using the quotient rule and trig identities i got there.Maybe could cancel out terms, but not sure.

Answer 45

Could you show me the rest of your work? Yes, you could go with cancellation.

Answer 46

Sure. Well we started with 1/csc(x)cot(x) like you suggested. Then i did the quotient rule.

\[(\csc(x)+\cot(x))(0)-(1)(-\csc(x)\cot(x)-\csc^2x) \]

Answer 47

Which equalled what i just posted above ^^

Answer 48

Because you divided by (g(x)^2

Answer 49

Yeah, good job. The rest?

Answer 50

I stopped there cause you wanted to see work.

Answer 51

Will be left with,
\[\frac{ \csc^2x }{ \csc(x)\cot(x) }\]

Answer 52

Final answer? Not really sure to be honest.

Answer 53

oops actually, the top will just be csc(x) not squared.

Answer 54

Well whatever thats the right answer, just checked. Thanks for the help.

Answer 55

Hmm. That'd give you \(\frac{1}{\cot(x)} = \tan(x)\)