Question

Simple derivative question, help?

Answer 1

\[y=\frac{ x^3\ln2x }{ e^xsinx }\]

Answer 2

you just want the answer, or what?

Answer 3

You first need to apply the quotient rule and then chain rule to both denominator and numerator..

Answer 4

where are you stuck?

Answer 5

I thought you weren't allowed to give out answers?
And I think I got it I just wanna see if I did it right

Answer 6

you can skip the quotient rule by rewriting as \[y = e^{-x}x^3 \csc x \ln 2x\]

Answer 7

Yeah you arent allowed to give out answers..
and yeah you can do that too.

Answer 8

No one gave out an answer, or even offered to give out an answer.

Answer 9

>.>
Well here's what I got:
\[\frac{ dy }{ dx }=\frac{ x^3\ln2x }{ e^xsinx }(\frac{ 3 }{ x }+\frac{ 1 }{ xln2 }-1-cotx)\]

Answer 10

That might not be correct...

Answer 11

Hm, the whole answer?

Answer 12

I'm not so sure about the dy/dx part ;-)

Answer 13

how did you come up with that \(\ln 2\)?

Answer 14

oops, that's suppose to be a xln2x

Answer 15

See, I told you it might not be correct :-)

Answer 16

I see >.>

Answer 17

I got it, haha thanks! @whpalmer4

Answer 18

what did you get for the final result?

Answer 19

ha.. actually i don't got it

Answer 20

\[\large y=\frac{ x^3\ln2x }{ e^xsinx }\]\[\large f'(x) = 3x^2(\ln(2x))+\left( \frac{x^3}{x}\right) = 3x^2\ln(2x) +x^2 \]\[\large g'(x)= e^x \sin(x)+e^xcos(x)\] that's a start.

Answer 21

i dont think you can differentiate numerator and the denominator separately?

Answer 22

I'm not.

Answer 23

Pfft, this girls got it :P

Answer 24

The quotient rule states \[y' = \frac{f'(x)\cdot g(x) -g'(x)\cdot f(x)}{g^2(x)}\]

Answer 25

So... i stated what the derivatives were, and you know what the other parts are, just....plug them in :)

Answer 26

oh, you were doing that.. sorry.

Answer 27

Ha, calm down Jhanny ^_^'

Answer 28

hes solving it :P

Answer 29

\[y = e^{-x}x^3\ln 2x \csc x\]let \(u=e^{-x}, v=x^3\ln 2x \csc x\) and apply the product rule:
\[\frac{dy}{dx} = x^3 \ln 2x \csc x \frac{d}{dx}[e^{-x}] + e^{-x}\frac{d}{dx}[x^3\csc x \ln 2x ]\]\[\frac{d}{dx}[e^{-x}] = -e^{-x}\]\[\frac{dy}{dx} =-e^{-x} x^3 \ln 2x \csc x + e^{-x}\frac{d}{dx}[x^3\csc x \ln 2x ]\]

Answer 30

Ugh this is too long!!

Answer 31

Sorry for a long problem :l

Answer 32

@Jhannybean I thought you loved long problems? :P

Answer 33

no <_< lol....

Answer 34

I'll upload a picture of what i've written. You can all experience my beautiful handwriting.

Answer 35

\[\frac{d}{dx}[x^3 \csc x\ln 2x] =\]Let \(u=x^3, v=\log 2x\csc x\)\[\frac{d}{dx}[x^3 \csc x\ln 2x] = \csc x\log 2x * 3x^2 + x^3\frac{d}{dx}[\csc x\log 2x]\]

\[\frac{d}{dx}[\csc x\log 2x] = \]
Let \(u=\csc x, v=\log 2x\)\[\frac{d}{dx}[\csc x\log2x] = \log 2x \frac{d}{dx}[\csc x]+\csc x\frac{d}{dx}[\log 2x]\] \[= -\cot x\csc x\log 2x + \csc x\frac{d}{dx}[\log 2x]\]

Answer 36

Night all. good Luck @Luigi0210 :)

Answer 37

\(u=2x\)\[\frac{d}{dx}[\ln 2x] = \frac{1}{ 2x}*2 = \frac{1}x\]

\[\frac{d}{dx}[x^3\csc x\ln 2x] =3x^2\csc x\ln 2x + x^3( -\cot x\csc x \log 2x + \frac{\csc x}{x})\]

\[\frac{dy}{dx} =-e^{-x} x^3 \ln 2x \csc x + e^{-x}\frac{d}{dx}[x^3\csc x \ln 2x ]\]
\[=-e^{-x} x^3 \ln 2x \csc x -e^{-x}(3x^2\csc x\ln 2x + x^3(-\cot x\csc x \ln 2x+\frac{\csc x}{x}))\]\[=-e^{-x}x^3\csc x\ln 2x - 3x^2e^{-x}\csc x\ln 2x+e^{-x}x^3\csc x\ln 2x\cot x-e^{-x}x^2\csc x\]\[=-e^{-x}x^2\csc x( x\ln 2x-3\ln 2x + x\cot\ln 2x-1)\]\[=-e^{-x}x^2\csc x(( x-3)\ln 2x-3x\ln 2x + x\cot\ln 2x-1)\]

Trying to spot mistakes looking at the LaTex code is painful!

Answer 38

first correction:
\[=-e^{-x}x^2\csc x(( x-3)\ln 2x-3x\ln 2x + x\cot x\ln 2x-1)\]