lilly21
Can someone check my calc prob and tell me if my answer is correct? Find the derivative of the function. y=Ine^x



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lilly21
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y(prime)=1/x(e^x)+e^x(In)
and my answer is y(prime)=e^x(1/x+In)

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

lilly21
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u lost me

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

lilly21
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well the problem has absolutely No parenthesis

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

suneja
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ya
is it 1

lilly21
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okay?

suneja
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is the correct ans 1

lilly21
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i have no idea this is an even problem and my book does not carry evens numbered problems

UnkleRhaukus
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\[y=\ln_ex\] ?

lilly21
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No simply lne^x

micahwood50
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lne^x = xlne = x
So y = x
Now derivative it.

lilly21
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y=In e^x

suneja
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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

lilly21
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you guys are completely confusing me

suneja
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did u get tis

micahwood50
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Do you know logarithm properties?

lilly21
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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?

suneja
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NO

lilly21
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i mean ln=1/x

micahwood50
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\[\ln a^b = b \ln a \]

suneja
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ya

micahwood50
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So \[\ln e^x = x \ln e\]

micahwood50
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And ln e = 1

micahwood50
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Do you know logarithm properties?

lilly21
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i did

suneja
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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

micahwood50
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Well, so y = x.
Now derivative it.

lilly21
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1

micahwood50
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Yeah. y' = 1

lilly21
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okay thanks

micahwood50
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No problem. Glad I helped you.

lilly21
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:]