Question

can someone walk me through the how to do derivatives of trig function

y= (sin x) (cosx)

Answer 1

dy/dx = (cosx) -(sinx) what I am missing here??

Answer 2

you should use product rule

Answer 3

\[(uv)' = uv' + vu'\]
remember that thingy?

Answer 4

okay so forgive me here I am rusty can you help me with this?

Answer 5

what you did was just \[u' + v'\]
you did not hold u or v constant... maybe a demo would be better

if i have x ln x the derivative of this is
\[x \frac{d}{dx} (\ln x) + \ln x \frac{dy}{dx} (x)\]
\[\implies x (\frac 1x) + \ln x(1)\]
\[\implies 1 + \ln x\]
does that help?

Answer 6

so if I understand with this rule you lower the poewer and times it for this

Answer 7

plus you have to consider a limt as goes to 0

Answer 8

no...it was just a coincidence....another example..
\[y = \sec x \tan x\]
\[\frac{dy}{dx} = \sec x\frac{d}{dx} (\tan x) + \tan x \frac{d}{dx} (\sec x)\]
\[\implies \sec x(\sec^2 x) + \tan x(\sec x \tan x)\]
\[\implies \sec^3 x + \sec x \tan^2 x\]

Answer 9

so in your problem
\[y = \sin x \cos x\]
it will be \[\implies \sin x \frac{d}{dx} (\cos x) + \cos x \frac{d}{dx} (\sin x)\]

Answer 10

so I need to keep at the start and end and make sure I use d/dx because the product rule calls for it y'(sin) = than sinx d/x (cosx) +cosx d/dx (sinx) due to the product rule is the way to solve this I am right here?

Answer 11

i think so? why dont you try and tell me your answer...then we can judge if you understood it

Answer 12

I got 'y(sin) is D/DX cosx -sinx + cosx

Answer 13

hmm no it seems you did not get it..

Answer 14

because I am d/Dx of sin(x) which is d/dx cos(x)

Answer 15

what?

Answer 16

I have a book that gives y' of trig functions that is where I am gettting this from. such what d/dx of sin(x) maybe I am confussed on that.

Answer 17

right from d/dx sin(x) says it's dy/dx is cos(x) is this right??

Answer 18

from what says what? im not getting what you're saying...

Answer 19

the derivative of sinx is cos x not dy/dx (cos x)

Answer 20

what is the derivate of sin(x)?

Answer 21

\[\frac{d}{dx} (\sin x) = \cos x \ne \frac{d}{dx} (\cos x)\]

Answer 22

\[\frac{d}{dx} (\cos x) = -\sin x \ne \frac{d}{dx} (\sin x)\]

Answer 23

is that better now?

Answer 24

okay so what is the key to this? product rule is one but how to do this so I can solve quickly in this case?

Answer 25

do i need flash cards and know all trig functions d/dx?

Answer 26

product rule is the only way....or if you know your trig identities well you know that
\[\sin x \cos x = \frac{\sin (2x)}{2}\]
derivative of that is \[\frac{\cos(2x) (2)}{2} \implies \cos (2x)\]
that's the shortcut but like i said you need to know the trig identities to recognize that...ad also it uses chain rule...a higher level of derivative than product rule

and in my opinion the trig function you just need toknow are sin x and cos x

Answer 27

what id are the ones I must know? I am in college calc 1 and I am trying to do this

Answer 28

haven't used calc 1 for a while I do understand it, I am still very rusty.

Answer 29

these are the derivatives my teacher taught me to memorize...just 5
\[\sin x \implies \cos x\]
\[\cos x \implies -\sin x\]
\[\ln x \implies \frac 1x\]
\[e^x \implies e^x\]
\[x^n \implies nx^{n-1}\]
and also the prduct, quotient and cchain rules

Answer 30

it seems like trig that have doesn't come to much in play here, I mean the unit circle and other stuff. I am guessing that I am only using 10 percent or less of trig in Calc 1 now? right?

Answer 31

what you just need in trig are the trig identities you need to know \[\sin^2 x + \cos^2 x = 1\]
\[\sin (2x) = 2\sin x \cos x\]
\[\cos (2x) = \cos^2 x - \sin^2 x\]
in my opinion those are te most important trig identities you'll use in calculus I

Answer 32

thanks any other tips here/ just asking??

Answer 33

dont memorize too many formulas or identitites...stick to the important ones and teach yourself how to obtain the other identities using basic identities...for example.. deriving tan x using sin x, cos x and quotient rule, etc.

Answer 34

do I need to take trig again or if I get what you are saying down will I be okay Calc 1?

Answer 35

you need algebra more than trig...especially in graphing...what im saying is important for 3/5 of differential calculus.. graphing is 2/5

Answer 36

and like i said what you need in trig are just the trig identities and i gave you the most commonly used identities... im pretty sure you wont encounter triangles in calculus I

Answer 37

y=f(x) what the graph is? or what is the graph y=f(x) +1