anonymous
  • anonymous
Can someone check my calc prob and tell me if my answer is correct? Find the derivative of the function. y=Ine^x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
y(prime)=1/x(e^x)+e^x(In) and my answer is y(prime)=e^x(1/x+In)
anonymous
  • anonymous
y=lne^x=e^x de^x/dx =e^x
anonymous
  • anonymous
u lost me

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More answers

anonymous
  • anonymous
hold the phone \(\ln(e^x)=x\)
anonymous
  • anonymous
well the problem has absolutely No parenthesis
anonymous
  • anonymous
sorry do u knw dat log n exp are mirror fns and that why they cancel out each other
anonymous
  • anonymous
ya is it 1
anonymous
  • anonymous
okay?
anonymous
  • anonymous
is the correct ans 1
anonymous
  • anonymous
i have no idea this is an even problem and my book does not carry evens numbered problems
UnkleRhaukus
  • UnkleRhaukus
\[y=\ln_ex\] ?
anonymous
  • anonymous
No simply lne^x
anonymous
  • anonymous
lne^x = xlne = x So y = x Now derivative it.
anonymous
  • anonymous
y=In e^x
anonymous
  • anonymous
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
anonymous
  • anonymous
you guys are completely confusing me
anonymous
  • anonymous
did u get tis
anonymous
  • anonymous
Do you know logarithm properties?
anonymous
  • anonymous
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?
anonymous
  • anonymous
NO
anonymous
  • anonymous
i mean ln=1/x
anonymous
  • anonymous
\[\ln a^b = b \ln a \]
anonymous
  • anonymous
ya
anonymous
  • anonymous
So \[\ln e^x = x \ln e\]
anonymous
  • anonymous
And ln e = 1
anonymous
  • anonymous
Do you know logarithm properties?
anonymous
  • anonymous
i did
anonymous
  • anonymous
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
anonymous
  • anonymous
Well, so y = x. Now derivative it.
anonymous
  • anonymous
1
anonymous
  • anonymous
Yeah. y' = 1
anonymous
  • anonymous
okay thanks
anonymous
  • anonymous
No problem. Glad I helped you.
anonymous
  • anonymous
:]

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