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The_Brain

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- Thread starter The_Brain
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The_Brain

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- #2

StephenPrivitera

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Take the natural log of both sides.

y=x^{x}

lny=xlnx

Differentiate with respect to x.

(1/y)(dy/dx)=lnx+x/x (chain rule, product rule)

dy/dx=y(lnx+1)

You could do the same thing with y=x^{a}, where a is a constant.

lny=alnx

dy/dx=y(a/x)=x^{a}(a/x)=ax^{a-1}

a formula I'm sure you know well

y=x

lny=xlnx

Differentiate with respect to x.

(1/y)(dy/dx)=lnx+x/x (chain rule, product rule)

dy/dx=y(lnx+1)

You could do the same thing with y=x

lny=alnx

dy/dx=y(a/x)=x

a formula I'm sure you know well

Last edited:

- #3

Guybrush Threepwood

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(e

- #4

check out the power rule, learn it , know it, LOVE IT! Its easy, i am just playing but check it out, ok. Any problems with it lemme know, k.Originally posted by The_Brain

Dx

- #5

quantumdude

Staff Emeritus

Science Advisor

Gold Member

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Originally posted by Dx

check out the power rule, learn it , know it, LOVE IT! Its easy, i am just playing but check it out, ok. Any problems with it lemme know, k.

No, the power rule is not applicable here because the exponent is not a constant. Stephen Privitera's solution is correct; go with that.

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