Question

find the derivative of cos2theta

Answer 1

can you please show the step-by-step process please? thankyou

Answer 2

First, tell me what shall be the derivative of cosx?

Answer 3

and you'll need chain rule, which i guess you already know

Answer 4

\[(-2)\sin (2\theta)\]

Answer 5

uhm, my answer for this is -4sintheta

Answer 6

yeah @nitz :)

Answer 7

Hi again @hartnn :)

Answer 8

hello :)

Answer 9

how did u get -4 ?

Answer 10

2 * -2 ?

Answer 11

here's the real question. \[\frac{ d }{ d \Theta } (\sin \Theta * \cos2\Theta) =\]

Answer 12

-2, yes
where does 2 come from ?

Answer 13

ohh

Answer 14

uhm, -sin2\[-\sin2\Theta * 2\] <<< chain rule?

Answer 15

that's a composite function, right? O_O

Answer 16

yes
so, derivative of cos 2theta = - sin2theta * derivative of 2 theta
= -2 sin 2 theta is correct :)

Answer 17

and whats your final answer for the actual question u posted?

Answer 18

im not sure actually. \[\cos \Theta*\cos2\Theta-4\sin \Theta*\sin \Theta\]

Answer 19

once again, where does that -4 come from :P

Answer 20

i keep on checking it on wolframalpha.com but i don't get the same answer :(

Answer 21

your first term is correct

Answer 22

uhm, come again please? where does my 4 come from? O_O

Answer 23

2nd term totally incorrect

Answer 24

im actually using product rule here. what should i use then?

Answer 25

for 2nd term
you'll keep sin theta as it as and derivate os 2theta

Answer 26

right ?

Answer 27

i meant derivate cos 2theta

Answer 28

\(\large \cos \theta \cos 2\theta + \sin \theta \dfrac{d}{dx}\cos 2\theta\)

you get this ?

Answer 29

i meant d/dtheta -_-

Answer 30

ohh. yeah, i get it

Answer 31

good, now we solved
derivative of cos 2theta = -2 sin 2 theta
just plug this in

Answer 32

\(\large \cos \theta \cos 2\theta + \sin \theta (-2\sin 2\theta)\)
simplify

Answer 33

oh. okay.

Answer 34

we can further simplify that?

Answer 35

but this, is something wrong here?

http://www.wolframalpha.com/input/?i=%28sin+theta+*+cos+%282theta%29%29+find+derivative

Answer 36

nothing much to simplify
cos t cos 2t -2 sin t sin 2t

wolf gave answer in terms of 3theta, do you require answer in terms of 3theta ?

Answer 37

else, cos t cos 2t -2 sin t sin 2t
can be totally acceptable final answer

Answer 38

oh. okay. thankyou :)

Answer 39

and can u pls help me with evaluating derivatives? :)

Answer 40

today i was going through this book..calculus for the clueless by bob millers part 1...perfect fit for me..you might try it out..not saying that you are clueless..but it's helpful

Answer 41

Hahaha. Sure thing x thanks

Answer 42

here are some questions @hartnn :)

Answer 43

\[2. g(s) = \sqrt[3]{s} + s^{2}; s'(3)=\]

Answer 44

\[3. h(t)=\sqrt{3t ^{2}-5t}; h' (\frac{ 1 }{ 6 }) = \]

Answer 45

looks like you should just find the respected derivatives and plug 3 on the first and 1/6 on the second into them

Answer 46

ohhh.

Answer 47

is that a typo on the first one? the function is g(s) but the second part is s'(3)

Answer 48

oh, i think so. but my worksheet says its 3. lol

its probably a typo ;) sorry

Answer 49

the 3 doestn bother me, its the fact that its wanting the derivative of the variable term. maybe someone else could make better sense of it than me. the second one is completely straightforward though

Answer 50

\[a. \frac{ d }{ dx }| _{x=1} (\sqrt{x ^{3}-1}) =\]
the subscript is x= 1

Answer 51

is this a new question?

Answer 52

\[b. \frac{ d }{ dx }| _{x=\frac{ 1 }{ 6 }} (3x ^{2}-4x+2) =\]

subscript is x=1/6

Answer 53

@lonnie455rich yep :)

Answer 54

\[c. \frac{ d }{ dr }| _{r=2} [(5r \times \cos \sqrt{r})]=\]

subscript is r=2

Answer 55

your subscripts look like your restrictions on x. I guess I don't understand the question?
are you just trying to find the derivatives?

Answer 56

uhm, it says here 'evaluate the derivatives at given points'

Answer 57

d. g(s)=s√3+s2; g′(3)=

Answer 58

e. g(s)=s√(3t^2 - 5t); h/(1/6)=

Answer 59

ahhh use the limit definition
(f(x+h)-f(x))/h

Answer 60

you may know h as delta x if that makes more sense to you

Answer 61

uhm, okay. imma try understand it :)

Answer 62

uhm, i dont get it. what will i use to substitute and stuff? :(

Answer 63

that should have been your first chapter in learning derivatives, no?
do b first since its easiest to check. then once you get the hand of it. the rest aren't so bad. here is an attachment of when I was learning this a few months ago

Answer 64

sorry, i've seen it in my notes here and im trying to remember it. lol

Answer 65

if its not mentioned to use limit definition of derivative, then we can use standard formulas

Answer 66

\( \large g(s) =\sqrt[3]s +s^2 = s^{1/3}+s^2\)
now we use the x^n formula of derivative, i think you know it, so i'll directly write next step
\(\large g'(s) = (1/3)s^{-2/3}+2s\)
to get g'(3), just plug in s= 3 in that and simplify :)

Answer 67

you wanted me to solve all 7 of them :O

Answer 68

ahh so if you have a point. for x you don't have to use the limit definition? just a nother form of f'(3) of f(x)=2x^2
I solved a lot of problems the hard way lol.

Answer 69

haha, as i said, use definition only if specifically mentioned :)

Answer 70

see, I just thought that if you to evaluate at a point you had to. its funny my intuition when I saw f'(3) was to compute the derivative and plug it in but when he said at a point I thought limit definition. frustrating sometimes. not knowing shortcuts.lol