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Mathematics 17 Online
OpenStudy (anonymous):

Where does the X come from?

OpenStudy (anonymous):

http://www.math.rutgers.edu/~sferry/MA135F14/135-f14/Fall2009ANS.pdf Number 9. Where does the X^8X come from

OpenStudy (anonymous):

in the answer

OpenStudy (anonymous):

y =x^(8x)

OpenStudy (anonymous):

I see that

OpenStudy (anonymous):

remember the power expansion log rule?

OpenStudy (anonymous):

in the first step he uses that to bring the 8x down in front on the left side

OpenStudy (anonymous):

yes i understand that

OpenStudy (anonymous):

dy/dx= x^n nx^(n-1)

OpenStudy (anonymous):

I just dont understand the second part

OpenStudy (anonymous):

I know derivative of y is 1/y dy/dx

OpenStudy (anonymous):

but how is the derivative of 8xlnx= 8lnx+8x*1/x

OpenStudy (anonymous):

ln y =ln (x^(8x))

OpenStudy (anonymous):

ah its the chain rule!

OpenStudy (anonymous):

then ln y =x8 ln (x)

OpenStudy (anonymous):

log rule

OpenStudy (anonymous):

oh the product rule what a fluttering idiot

OpenStudy (anonymous):

product rule yea not chaing rule lol

OpenStudy (anonymous):

i am talking about how stupid i am

OpenStudy (anonymous):

so how does x^8x come back in the end

OpenStudy (anonymous):

he/she use the log role to solve the problem... 1- take the ln for both side... ln y =ln (x^(8x)) 2-

OpenStudy (anonymous):

wait

OpenStudy (anonymous):

I understand all the steps except the end

OpenStudy (anonymous):

I just dont understand the last step

OpenStudy (anonymous):

2- rewrite the eq. using log rule ln y= 8x ln x

OpenStudy (anonymous):

last step puting y value

OpenStudy (anonymous):

you're really not answering my question

OpenStudy (anonymous):

I am asking how did the x^8x appear at the end

OpenStudy (anonymous):

dy/dx (y)= 1/y y' .....

OpenStudy (anonymous):

yes i understand that step

OpenStudy (anonymous):

the y value...

OpenStudy (anonymous):

goes to the right side i understand but in the last step it disappeared

OpenStudy (anonymous):

the Y disappeared at the end

OpenStudy (anonymous):

it goes y'=x^8x(8lnx+8)

OpenStudy (anonymous):

yes putting the value of it which is x^8x

OpenStudy (anonymous):

I finally understand

OpenStudy (anonymous):

Thank you

OpenStudy (anonymous):

YW

OpenStudy (anonymous):

OH DUUDE, because its earlier defined as y literally equalling x^8x, you can substitute it when you move the 1/y over

OpenStudy (anonymous):

I was trying to understand that since the beginning

OpenStudy (anonymous):

i always thought of that as the last thing in my classes lol

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

i thought its history of math question -.-

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