Question

@jim_thompson5910

Once again, waiting for file :P

Answer 1

You forgot to apply chain rule for cos2x

Answer 2

she factored out that 2

Answer 3

it's correct

Answer 4

Yeah

Answer 5

I'm talking about line 2

And how did you get from \(2x\sin(2x)\) to \(\sin(4x)\) at line 3?

Answer 6

It's supposed to be:

sin4x+(x^2cos2x)(2x)

Then I factored out the 2x.

I just forgot to write it because I was moving fast

Answer 7

Wait no I dont need that.

I didn't do the chin rule

I did the product rule?

Answer 8

hmm, still don't see where you get sin 4x from.

don't matter, you managed to get correct answer.

Answer 9

yeah I agree, the sin(4x) doesn't fit

Answer 10

f'(x)(g(x))+f(x)(g'(x))

Answer 11

(x^2)(sin 2x)

deriv of x^2 is 2x
deriv of sin 2x is cos 2x
2x(sin2x)+(x^2)(cos 2x)

Answer 12

I guess 2x(sin2x) is not sin 4x

Answer 13

well, you forgot to apply chain rule \(\dfrac{d}{dx} sin(2x) = 2\cos(2x)\)

Answer 14

I don't understand why I need to if I'm just doing the product rule

Answer 15

well yes, you did product rule correctly, but whiling taking derivative of sin2x, you need to apply chain rule there

Answer 16

So is the 2nd line supposed to be:

2x(sin2x)+x^2(cos2x)(2x) ???

Answer 17

Or just (2) at the end?

Answer 18

just 2 since the derivative of 2x is 2

Answer 19

Chain rule: \(\dfrac{d}{dx}f(g(x)) = f'(g(x))g'(x)\)
In this case, f(g(x)) is sin2x, where f(x) is sinx and g(x) is 2x

Answer 20

I thought so but I wasn't sure if I was using the derivative of 2x or x^2

Yeah I understand now because if I jsut have take the derivative of sin2x

u would =2x and it would be cos2x(2) would be the derivative.

Answer 21

yep

Answer 22

Thanks

Answer 23

our pleasure

Answer 24

Okay @jim_thompson5910

I know I have to do some table chart thing but I dont remember how lol

An no I'm not allowed to just graph it and look, it says I cant.

Answer 25

My file isn't uploading so I'll write it out

Answer 26

Let f be the function with derivative f'(x)= \[f'(x)=x^2-\frac{ 2 }{ x }\]

On which interval is x decreasing?

I found the critical numbers and they are x= +-rad2 right?

Then the top part of the table is like

-infinity<=x<=-rad2 then
-rad2<=x<=rad2 then
rad2<=x<=infinity

?

Answer 27

Oh I think it just sent to me so I'll upload a file now

Answer 28

when you multiplied both sides by x, you should have gotten this

\[\Large x^3 - 2 = 0\]

Answer 29

Why...?

Answer 30

x^2+2/x=0

0*x is 0

Answer 31

Sorry -2 not +

Answer 32

I multiplied the 2/x by x and 0 by x to get rid of the x on the left side

Answer 33

you have to do that to every term though

Answer 34

Ohh okay

Answer 35

so things are balanced out

Answer 36

So that changes all the rad2 to cube root 2s

Answer 37

Then I pick values and plug them in to see if they are positive of - right?

Answer 38

+ or - *

Answer 39

I'll use -2,0,and 2

Answer 40

Wait is 0 a critical number too???

Answer 41

no only \[\Large x = \sqrt[3]{2}\]

Answer 42

this is the only root of f'(x)

Answer 43

Okay.. I plug my values into x^2-2/x right?

Answer 44

I'm not getting any negative numbers to show it's decreasing

Answer 45

hmm you're using -2, 0, 2 right?

Answer 46

oh i see why

Answer 47

those x values do produce y values above the x axis

Answer 48

you were right in saying that x = 0 is like a boundary point

so pick a point between 0 and cuberoot(2)

Answer 49

Oh? I dont understand... lol

WHy is x a boundary point exactly?

Answer 50

because you can't divide by zero

Answer 51

So is 0 a critical point?

Answer 52

Well not a critical point but it does in the table

Answer 53

I think what goes in the table is the critical numbers and what x cannot equal right?

Answer 54

f(x) is decreasing \[[0,\sqrt[3]{2}]\]

Answer 55

that should be a ( in front of zero sorry

Answer 56

sry got distracted for a bit

Answer 57

It's okay

Answer 58

0 isn't a critical point because it's not even in the domain

Answer 59

I know but x cant =0.

And I remember my teacher saying to include that in the table

Answer 60

and good, it's decreasing on the interval \[(0,\sqrt[3]{2})\]

Answer 61

Just like if it was like x/x+2 you'd have your critical point in the table, but also -2 in the table

Answer 62

maybe he was referring to a problem where x = 0 is defined?

Answer 63

oh as like a boundary point, yeah I guess you could do that

Answer 64

Yeah thats what he says to do

Answer 65

then yeah, also \[(0,\sqrt[3]{2})\] is correct

Answer 66

Is this right because I'm not 100% convinced in myself

Answer 67

You didn't need to find the equation of the tangent line

You just needed to find the slope of the tangent line at x = 1

Answer 68

But it's still good practice

Answer 69

Oh... how would I have done that?

Answer 70

you could have stopped at the slope formula where you got the result of 3 (the first time)

Answer 71

Oh so all had to do was the slope formula? lol

Answer 72

yep since derivatives visually represent slopes

Answer 73

slopes of tangent lines

Answer 74

Ahhh right

Answer 75

Okay so I know to find concavity I have to find the 2nd derivative and make that table again.

Did I do the 2 derivatives right?

Answer 76

Welll my file isn't uploading so:

\[f(x)=2xe^x\]

Answer 77

I got by doing the product rule 2x(e^x)

\[f'(x)=2e^x+2xe^x\]

Answer 78

Then I did the product rule with 2(e^x) and added it to the product rule of 2x(e^x) \[f''(x)=2e^x+2e^x+2xe^x\]

Answer 79

you don't need to do the product rule with 2(e^x)

you can use the constant rule

Answer 80

or constant multiple rule

Answer 81

What's that

Answer 82

if y = k*f(x)

then y' = k*f ' (x)

Answer 83

k is a constant

Answer 84

Oh like 2e^x^1 would be 2*1e^x

Answer 85

Like the power rule?

Answer 86

sorta,

2e^x = 2*e^x

ignore the constant and derive e^x to get e^x, then stick the constant back on

----------

so the derivative of 2e^x is 2e^x

Answer 87

Okay but doing the power rule helps me visualize and I still get the right answer.

I have the right answer of the 2nd derivative right?

Answer 88

anyways, you'll get

\[\Large f''(x)=4e^x+2xe^x\]

after you combine like terms

Answer 89

now find the root of f''(x)

Answer 90

Yay I did it right

Answer 91

Dont you take the ln of both sides to get

4+2x=1

Answer 92

Wait is the ln(0)=1 or ln(0)=0?

Answer 93

Is the critical # -2?

Answer 94

you set it to zero, then solve for x

Answer 95

I know.. but then you have to get rid of the e^xs so you take the ln to cancel them right?