Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (mtalhahassan2):

Differentiate y= e^x / x

OpenStudy (baru):

y=\(x^{-1} \times e^x \) use chain rule

OpenStudy (mtalhahassan2):

[img]/drawings/45bfaeb002eb571b3a8c239fe944391d.png[/img]

OpenStudy (mtalhahassan2):

y= e^x (x^-1) -1 e^x (x^-2)

OpenStudy (mtalhahassan2):

right??

OpenStudy (baru):

\(y= (x^{-1}e^x) + (-x^{-2}e^x)\)

OpenStudy (baru):

is that what you got?

OpenStudy (mtalhahassan2):

yes i got the same thing but are you not missing -1

OpenStudy (baru):

see the second term: i have written (- x)

OpenStudy (mtalhahassan2):

oh ok so you put the -1 inside the backet

OpenStudy (baru):

yes

OpenStudy (mtalhahassan2):

wait at the back of the book they have different answer i guess they simplify further

OpenStudy (baru):

[y=\frac{e^x}{x}-\frac{e^x}{x^2} \]

imqwerty (imqwerty):

lol finally!!! some life detected on os

OpenStudy (baru):

\[y=\frac{e^x}{x}-\frac{e^x}{x^2} \]

OpenStudy (baru):

just take LCM

OpenStudy (baru):

yeah, things have become really quiet qwerty

imqwerty (imqwerty):

yes its too quiet :) i do research wrk on os when its quiet like this

OpenStudy (baru):

lol...research?? i'm suspicious O.o

OpenStudy (baru):

@MTALHAHASSAN2 your answer matched with textbook?

imqwerty (imqwerty):

yes my current project is to make a blank username..i can make it tho but only if the character->"%" was allowed ): but its not m wrkin on finding ip's of diff ppl e.e i've found 10so far

OpenStudy (mtalhahassan2):

no they are getting like y1= e^x / x^2 (x-1)

OpenStudy (baru):

like i said, just take common denominator \[y=\frac{xe^x}{x^2}-\frac{e^x}{x^2} \]

OpenStudy (baru):

take e^x/x^2 common

OpenStudy (mtalhahassan2):

i goted that but how they are getting (x-1)

OpenStudy (baru):

taking \(e^x/x^2\) common \[y=\frac{e^x}{x^2}(x-1)\]

OpenStudy (baru):

@MTALHAHASSAN2

OpenStudy (mtalhahassan2):

So this is what we got right?? y= e^x (x^-1) + e^x (-x^-2)

OpenStudy (mtalhahassan2):

e^x be the commom

OpenStudy (baru):

yes

OpenStudy (mtalhahassan2):

I am confuse

OpenStudy (mtalhahassan2):

know how can we solve

OpenStudy (baru):

you know this right: \(x^{-1}=\frac{1}{x}\)

OpenStudy (mtalhahassan2):

yes

OpenStudy (baru):

wait website hanging

OpenStudy (mtalhahassan2):

can you help me with y= log 10 (1-x+x^3)

OpenStudy (astrophysics):

Hint:\[\frac{ d }{ dx }\log_ax =\frac{ 1 }{ xlna }\]

OpenStudy (unklerhaukus):

\[y= \frac{e^x}x\] With the quotient rule \[y' =\frac{xe^x-e^x}{x^2}\] factoring the numerator \[=\frac{(x-1)e^x}{x^2}\]

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!