Question

Can someone check my calc prob and tell me if my answer is correct? Find the derivative of the function. y=Ine^x

Answer 1

y(prime)=1/x(e^x)+e^x(In)
and my answer is y(prime)=e^x(1/x+In)

Answer 2

y=lne^x=e^x
de^x/dx =e^x

Answer 3

u lost me

Answer 4

hold the phone
\(\ln(e^x)=x\)

Answer 5

well the problem has absolutely No parenthesis

Answer 6

sorry
do u knw dat log n exp are mirror fns and that why they cancel out each other

Answer 7

ya
is it 1

Answer 8

okay?

Answer 9

is the correct ans 1

Answer 10

i have no idea this is an even problem and my book does not carry evens numbered problems

Answer 11

\[y=\ln_ex\] ?

Answer 12

No simply lne^x

Answer 13

lne^x = xlne = x

So y = x

Now derivative it.

Answer 14

y=In e^x

Answer 15

see ur que was to diff y= lne^x
now as i said ln and exp cancels out each other
so lne^x=x
ie y=x
dx/dx=1

Answer 16

you guys are completely confusing me

Answer 17

did u get tis

Answer 18

Do you know logarithm properties?

Answer 19

one question, is it possible to cancel well when rewritten.....ln=1/w so we have 1/x e^x is it possible to cancel out the x's?

Answer 20

NO

Answer 21

i mean ln=1/x

Answer 22

\[\ln a^b = b \ln a \]

Answer 23

ya

Answer 24

So \[\ln e^x = x \ln e\]

Answer 25

And ln e = 1

Answer 26

Do you know logarithm properties?

Answer 27

i did

Answer 28

you can do that way too
but rem tis log n exponential r mirrors function dats y can cancel them out
and wat micahwood50 says is also correct

Answer 29

Well, so y = x.

Now derivative it.

Answer 30

1

Answer 31

Yeah. y' = 1

Answer 32

okay thanks

Answer 33

No problem. Glad I helped you.

Answer 34

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