EmLo12
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Why isn't y = lne^x so dy/dx = 1/e^x correct?
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Eulogy
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(Original post by EmLo12)
Why isn't y = lne^x so dy/dx = 1/e^x correct?
ln(e^x) is just x, if I understand correctly?
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NishatM
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y = lne^x is just y = x isn't it? meaning dy/dx would be 1
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_gcx
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(Original post by EmLo12)
Why isn't y = lne^x so dy/dx = 1/e^x correct?
No, you're misquoting the chain rule. Using the chain rule,

\displaystyle \frac {\mathrm d} {\mathrm d x} \ln e^x = \frac {\mathrm d} {\mathrm d(e^x)} \ln e^x \cdot \frac {\mathrm d} {\mathrm d x} e^x = \frac 1 {e^x} \cdot e^x = 1

Which is what you'd get from recognising that \displaystyle \ln e^x = x \ln e = x.
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MR1999
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(Original post by EmLo12)
Why isn't y = lne^x so dy/dx = 1/e^x correct?
The chain rule applied to  \ln f(x) shows that

 \displaystyle \frac{d}{dx} (\ln f(x)) = \frac{f'(x)}{f(x)}

Applying this to \displaystyle \ln e^x will give you your answer, although simplifying the expression is much easier.
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