Question

Dany612

Answer 1

present!

Answer 2

\[\sqrt{2x-3}\]

Answer 3

this is the problem?

Answer 4

you forgot the outside x

Answer 5

oh ok.. \[x \sqrt{2x-3}\]

Answer 6

yes

Answer 7

We need product rule and chain rule here.

Answer 8

\[(x)(\sqrt{2x-3})\prime + (1)(\sqrt{2x-3})\]

Answer 9

Ok that is product rule.. now we need chain rule to find the derivative of \[\sqrt{2x-3}\]

Answer 10

f(x) = sqrt(x)
g(x) = 2x-3

Answer 11

so the derivative of \[\sqrt{2x-3}\] is \[\frac{1}{2\sqrt{2x-3}}* 2\]

Answer 12

Which is \[\frac{2}{2\sqrt{2x-3}}\] or \[\frac{1}{\sqrt{2x-3}}\] because the 2's cancel

Answer 13

so now the derivative will look like combining the chain and product rule.\[(x)(\frac{1}{\sqrt{2x-3}}) + \sqrt{2x-3}\] or \[\frac{x}{\sqrt{2x-3}}+ \sqrt{2x-3}\]

Answer 14

You can leave it like this or find a common denominator.

Answer 15

okay

Answer 16

Finding a common denominator is probably best so the common denominator is \[\sqrt{2x-3}\] so you do \[\frac{x}{\sqrt{2x-3}} + \frac{\sqrt{2x-3}}{1}*\frac{\sqrt{2x-3}}{\sqrt{2x-3}}\]

Answer 17

so you are left with \[\frac{x + 2x-3}{\sqrt{2x-3}}\] or \[\frac{3x - 3}{\sqrt{2x-3}}\]you can factor out a 3 in the numerator so the final answer is \[\frac{3(x-1)}{\sqrt{2x-3}}\]

Answer 18

Anything you dont understand?

Answer 19

not yet

Answer 20

Cool.

Answer 21

okay I know I should know this but the multiplying by the donimator, i need more elaboration on that.

Answer 22

ok you have \[\frac{x}{\sqrt{2x-3}} + \frac{\sqrt{2x-3}}{1}\]

Answer 23

in order to add fractions you need a common denominator.

Answer 24

Common denominator in this case is \[(\sqrt{2x-3})(1) = \sqrt{2x-3}\]

Answer 25

so you multiply any fraction that doesnt have \[\sqrt{2x-3}\] in the denominator.

Answer 26

By \[\frac{\sqrt{2x-3}}{\sqrt{2x-3}}\]

Answer 27

oh wait nvm, i got it

Answer 28

The first fraction already has \[\sqrt{2x-3}\] in the denominator so no need to multiply anything to it.. The second fraction on the other hand has a 1 in the denominator.. But we need sqrt(2x-3) in the denominator.. so we multiply the second fraction by
(sqrt(2x-3))/(sqrt(2x-3)).. like this \[\frac{x}{\sqrt{2x-3}} + \frac{\sqrt{2x-3}}{1}*\frac{\sqrt{2x-3}}{\sqrt{2x-3}}\]

Answer 29

You understand now??

Answer 30

Yeah I did, sorry for making you right that whole thing.

Answer 31

Its alright.

Answer 32

okay so thats all the hlep I needed. Thanks!

Answer 33

No problem.. Im suprised i got it right due to the fact that im tired lol..

Answer 34

lol how come?

Answer 35

Lol i have no idea.. I havent done much today lol.. Just played MW3

Answer 36

lol I bet its all that hard concentration on killing enemies hah

Answer 37

lol or raging one of the two haha

Answer 38

I wonder if I'll enjoy those benifits lol because I haven't played MW3 yet or have plans to buy it.

Answer 39

Lol its pretty fun, i like the first person shooter games though.. even though im not all that good at them haha

Answer 40

Yeah same here, but I fear of being addicted.

Answer 41

do you mind lending a hand on another problem just like this one. for some reason I got it wrong.

Answer 42

lol i know!!! Im going to have to resist playing so i can do my hw lol

Answer 43

and yeah i can help

Answer 44

x(1-2x)^3

Answer 45

i wonder where i messed up

Answer 46

Ok we need product rule and chain rule again.

Answer 47

Product rule:\[(x)((1-2x)^3)\prime + (1-2x)^3\]

Answer 48

ok so far so good

Answer 49

i skipped the derivative of "x" on the right side of the plus sign because it is 1.

Answer 50

And something X 1 is something.

Answer 51

ok now we need chain rule for \[(1-2x)^3\]

Answer 52

f(x) = x^3
g(x) = 1-2x

So the derivative is\[2(1-2x)^2*(-2)\]

Answer 53

wait why is the coefficient 2 instead of 3?

Answer 54

Now combine chain rule and product rule\[x*2(1-2x)^2*(-2) + (1-2x)^3 \]

Answer 55

Because you have to use power rule.

Answer 56

power rule says \[f(x) = x^{n}\] then \[f \prime (x) = nx^{n-1}\]

Answer 57

i know but x^3 is 3x^2 so 3(1-2x)^2??

Answer 58

you dont have the same n

Answer 59

oh i made a type-o

Answer 60

no wonder you're confused lol

Answer 61

yeah I know my product rule lol

Answer 62

it should be \[x * 3(1-2x)^2 * (-2) + (1-2x)^3\]

Answer 63

okay thats exactly what I got, now to simplify

Answer 64

ok we can factor out at \[(1-2x)^2\]

Answer 65

\[(1-2x)^2[x * 3 + 1 - 2x]\]

Answer 66

so the derivative should be \[(1-2x)^2[x + 1]\]

Answer 67

let me see if that is correct.

Answer 68

hold on what happened to your -2?

Answer 69

i have -6x(1-2x)^2+(1-2x)^3

Answer 70

And i forgot the -2

Answer 71

see im tired lol.. let me fix it lol

Answer 72

\[(1-2x)^2[-6x + 1 - 2x]\]

Answer 73

yes thats what i got

Answer 74

which is \[(1-2x)^2[-8x + 1]\] then factor out a -1 so the x term in the parenthesis is positive.. \[-(1-2x)^2[8x-1]\]

Answer 75

And that is correct.. woohoo xD

Answer 76

yes, I see what my problem was. I forgot to add join the commons together which were -6 and -2 xD omg I'm such a fail.

Answer 77

-6x and -2x*

Answer 78

Ahh so you did it right but added wrong.. i do that too so dont feel bad lol

Answer 79

haha yeah simple mistkes. well thanks Tyler! Your awesome and a life saver!. sorry for bothering you when you felt lazy lol

Answer 80

i mean tired

Answer 81

lol not a problem!! If you have anymore questions dont be afraid to ask!!

Answer 82

no thats it. I finished up my 20 hw problems. I do have another question but ill ask you on facebook bec i gotta switch computers.

Answer 83

oh ok

Answer 84

wait are u still here, i just switched computers

Answer 85

yeah

Answer 86

so your studying computer engineering right?

Answer 87

yeah