Question

How do you derive this velocity function into acceleration? y=-50cos(x)sin(25x)-2sin(x)cos(25x)

Answer 1

preferably using the product rule

Answer 2

It doesn't look like you have much of a choice but to use the product rule :)
But can you also use chain rule?

Answer 3

yeah we can use whatever rule we like i guess :) i just dont even know how to do it..

Answer 4

Chain rule, though. It's kind of the key here. Can you use it?

Answer 5

yeah i can, but i dont know where to use it exactly?

Answer 6

Okay... first, we clarify a few things...if y is the velocity function, then the acceleration is the derivative of the velocity function...\[\large a=\frac{d}{dx}\left( -50\cos(x)\sin(25x)-2\sin(x)\cos(25x) \right)\]

Answer 7

yep, i get that :)

Answer 8

what do you mean by sin(25x)????

Answer 9

Well, you know that the differentiation operator distributes over addition (subtraction) and that constants can just be moved out (in other words, it's a linear operator)
So...

Answer 10

uh, -50cos(x)*sin(25x)

Answer 11

\[\large a= -50\frac{d}{dx}(\cos(x)\sin(25x))-2\frac{d}{dx}(\sin(x)\cos(25x)) \]

Answer 12

I have a question, the derivative of sinx is cosx, so is the derivative for sin(25x) = cos(25x)

Answer 13

is 25cos25x

Answer 14

whatever is in front of the x value comes out the front also. Im still a little condfused?

Answer 15

Let's focus on this part
\[\large \frac{d}{dx}(\cos(x)\sin(25x))\]Product rule... can you state it?

Answer 16

y=uv'+u'v

Answer 17

there is nothing called as such, cos(25x) means the angle is only multiplied by 25, however 25cos(x) means that 25 is multiplied by the cos value of x

Answer 18

what about the constant, -50?

Answer 19

woah, what? we were taught that whatever value that is in front of x, comes out the front and multiplies with any constants?

Answer 20

The constant was taken out of the differentiation operator. You know that
\[\large \frac{d}{dx}cf(x)=c\frac{d}{dx}f(x)\]

Answer 21

no, whats that?

Answer 22

Constants may move in and out of derivatives :D

Answer 23

aha okay :)

Answer 24

That is to say, the derivative of a constant times the function is the constant times the derivative of the function.

Answer 25

True.

Answer 26

ah okay, so the -50 stays out the front

Answer 27

We can prove that... using the product rule...
\[\large \frac{d}{dx}cf(x)=c\frac{d}{dx}f(x)+f(x)\frac{d}{dx}c\]Derivative of a constant is zero...
\[\large \frac{d}{dx}cf(x)=c\frac{d}{dx}f(x)+f(x)\cdot0=c\frac{d}{dx}f(x)\]done. Shall we continue?

Answer 28

sure :)

Answer 29

Back to this...
\[\large \frac{d}{dx}(\cos(x)\sin(25x))\]
You can let u = cos x
and v = sin(25x)
Now apply the product rule...

Answer 30

oh, so you have to do the product rule for each side?

Answer 31

What do you mean by "each side" The other side (left) only has a...

Answer 32

Or did you mean for each term? In that case, yes.

Answer 33

yeah term. I have a question. -50cos(x) derives to be -50(-sinx), but do the negatives cancel?

Answer 34

Of course they do :)
The things you learned in algebra still apply... very much :D

Answer 35

alright, so u'=50sin(x) and v'=-2cos(x)*25-sin(25x), so 50cos(x)sin(25x)

Answer 36

y=-1250*(cos x)*cos (25*x) + 50*(sin (25*x))*sin x - (-50*(sin x)*sin (25*x) + 2*(cos (25*x))*cos x)

that is the derivative from graphmatica

Answer 37

Nope... the derivative of a product is NOT the product of their derivatives...
(uv)' = uv' + u'v
You stated this... remember? :)

Answer 38

so what did i do wrong?

Answer 39

oh wait, i was doing the other terms!

Answer 40

You made it so that
(uv)' = u'v'

Answer 41

too. So careful

Answer 42

u'=50sin(x) and v'=25cos(x)

Answer 43

Okay. Now product rule. And be careful ;)

Answer 44

Wait... v'
Redo it.

Answer 45

y'=-1250cos(x)cos(25x)+50sin(x)sin(25x)

Answer 46

yeah i fixed v' up, my mistake!

Answer 47

Wait....patience... let's start over...
or at least... why is it 1250?

Answer 48

constants multiplied. -50cos x 25cos.. -50x25=-1250

Answer 49

that side is right according to graphmatica :)

Answer 50

But... something's wrong... we already brought out the -50. Why did you include it in your u?

Answer 51

cause i hadnt bought it in anywhere. I included it with my u because i let u=-50cos(x)

Answer 52

Okay, I've seen my error. Sorry, my bad. Please continue.

Answer 53

when you do the product rule for the second term, what do you do with them all at the end? and thats alright!

Answer 54

You add them.

Answer 55

Keeping in mind that there was a minus sign in between them, of course.

Answer 56

Remember...\[\large a= -50\frac{d}{dx}(\cos(x)\sin(25x))-2\frac{d}{dx}(\sin(x)\cos(25x))\]And you've already done the first product rule bit...

Answer 57

so when i do the second product rule bit, i just add them all other than the minus?

Answer 58

Well, distribute the minus sign. Differentiate the second term, first. I can't put in words what to do, I'd rather demonstrate, to avoid confusion.

Answer 59

hold on, would you be able to check my second product rule part?

Answer 60

Sure

Answer 61

alright. i let u=-2sin(x), so u'=-2cos(x)
i let v = cos(25x), so v'=-25sin(25x). When i used the product rule, i got:
50sin(x)sin(25x) - 2xos(x)cos(25x)

Answer 62

Nicely done :)
Now the only thing left to do is add them up :D

Answer 63

i dont think im going to get the same as the graphing program though..

Answer 64

I know it's daunting to check it, but actually, when you add them up, you DO get the same answer :)

Answer 65

so it would be:
-1250cos(x)cos(25x)+50sin(x)sin(25x)-50sin(x)sin(25x)+2cos(x)cos(25x)

Answer 66

Copying your post...
y=-1250*(cos x)*cos (25*x) + 50*(sin (25*x))*sin x - (-50*(sin x)*sin (25*x) + 2*(cos (25*x))*cos x)

Answer 67

for the second half, it says that it is meant to be -50(-sin...) but, considering that in the product rule it was -2sin(x)-25sin(25x), wouldnt the negatives cancel?

Answer 68

And uhh, you got a sign wrong...

Answer 69

and the minus sign would just make it -50

Answer 70

In this post...


so it would be:
-1250cos(x)cos(25x)+50sin(x)sin(25x)-50sin(x)sin(25x)+2cos(x)cos(25x)
You got a sign wrong right.............................................^ here

Answer 71

im confused about the sign too

Answer 72

Because you also posted...
alright. i let u=-2sin(x), so u'=-2cos(x)
i let v = cos(25x), so v'=-25sin(25x). When i used the product rule, i got:
50sin(x)sin(25x) - 2xos(x)cos(25x)
See this sign ......^ :)

Answer 73

yeah makes sense! why is it + in the answer? and why is it -50(-sin..)

Answer 74

Oh... I see. You distributed the minus sign. But that wasn't necessary, as you already considered the minus sign when you let u = -2sin(x)
Signs are VERY confusing, so be EXTRA careful, ok?

Answer 75

so what have i done wrong? im so confused now :(

Answer 76

also, can you have y''=? cause its derived twice

Answer 77

What you did was you distributed the minus sign like this
-(50sin(x)sin(25x)-2cos(x)cos(25x))When you shouldn't have done so, because, there is no longer a minus sign there, as you have already considered that minus sign when you let u = -2sin(x)

Answer 78

and i cant have acceleration, cause there is no time!

Answer 79

x is time.

Answer 80

At least, so I assumed.

Answer 81

a(x) then? and how do i change it so its right?

Answer 82

Again, just add them up, without frills.

Answer 83

Because this has been so long, I'll repost your posts..

y'=-1250cos(x)cos(25x)+50sin(x)sin(25x)

alright. i let u=-2sin(x), so u'=-2cos(x)
i let v = cos(25x), so v'=-25sin(25x). When i used the product rule, i got:
50sin(x)sin(25x) - 2xos(x)cos(25x)

Answer 84

so the second half of the product rule would be -50(-sinx)sin(25x)-2cos(x)cos(25)

Answer 85

Yeah. And add them up.

Answer 86

but why is the sin still negative if it cancels?

Answer 87

It isn't.

Answer 88

-2sin(x)-25sin(x)-2cos(x)cos(25x)
would be: 50sin(x)sin(25x)-2cos(x)cos(25x) cause the -2 and 025 make positive 50. But when you combine it all, the minus goes before the 50, but why does the sin become negative again?

Answer 89

The first time you used the product rule on the second term, you got it right, but then, you incorrectly took its negative when you shouldn't have done so...
Posting it again...

alright. i let u=-2sin(x), so u'=-2cos(x)
i let v = cos(25x), so v'=-25sin(25x). When i used the product rule, i got:
50sin(x)sin(25x) - 2xos(x)cos(25x)

Answer 90

y=-1250*(cos x)*cos (25*x) + 50*(sin (25*x))*sin x - (-50*(sin x)*sin (25*x) + 2*(cos (25*x))*cos x)
....................................................................................^

Answer 91

sin became negative because the derivative of cos x is -sin x
Also, it's
50sin(x)sin(25x) by the way

Answer 92

convert the cos*cos and sin*"sin products to sums
your life'd get much easier

Answer 93

-2sin(x)*-25sin(25x) = 50sin(x)sin(25x) am i right?

Answer 94

That's right.

Answer 95

\[\cos(x)\sin(25x)={1\over2}[\sin(x+25x)+sin(x-25x)]\\
\sin(x)\cos(25x)={1\over2}[\sin(x+25x)-\sin(x-25x)]
\]

Answer 96

all you are left with is buch of sines adding

Answer 97

si why is it that when you combine and add them all together, it goes - 50 (-sin..)
where did the (-sin..) come from if the negatives cancelled to make a positive!

Answer 98

Where the -sin came from? Because the derivative of cos x is -sin x