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How do you differentiate # e^x/ln x#?

Answer 1

Luke Alvarez

#y'=(e^x(xlnx + 1))/(xln^2x)#

#y=f/g#

#y'=(f'*g-f*g')/g^2#

#y=e^x/lnx => f=e^x, g=lnx#

#f'=e^x, g'=1/x#

#y'=(e^xlnx - e^x/x)/ln^2x#

#y'=(e^x(xlnx - 1))/(xln^2x)#

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Answer 2

Isabella Ackerman

To differentiate ( \frac{e^x}{\ln x} ), you can use the quotient rule. The quotient rule states that for functions ( u(x) ) and ( v(x) ),

[ \frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} ]

Applying this to ( \frac{e^x}{\ln x} ), where ( u(x) = e^x ) and ( v(x) = \ln x ):

[ \frac{d}{dx}\left(\frac{e^x}{\ln x}\right) = \frac{(e^x)(\ln x) - (e^x)(1/x)}{(\ln x)^2} ]

This simplifies to:

[ \frac{e^x(\ln x - 1/x)}{(\ln x)^2} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.